Reduction for a Terminating Bivariate Hypergeometric Appell Series ℱ₁ (II)

Abstract

This paper studies a terminating (“modified”) Appell function ${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y),$ with $\alpha \in {\mathbb{Z}}_{\ge 1}$ and $\beta ,{\beta }^{\prime }\in {\mathbb{Z}}_{\le -1}$, together with the associated terminating Gauss function ${}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};z\right).$ The reduction formula for the Appell function ${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)$ to ${\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{1-y}\right)}$ breaks down when $(\beta ,{\beta }^{\prime })$ are integers less than or equal to $-1$ and it needs to be substituted with a revised identity that includes an explicit additional correction term ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$. This correction term is initially computed for specific cases (particularly for $\alpha =1,2,3$) and subsequently formulated and verified generally through mathematical induction on $\alpha$. The final expression demonstrates a structured pattern reminiscent of binomial/Pascal coefficients and leads to various corollaries, including simplified boundary scenarios (such as when $\alpha =1$).

1. Introduction

In [1], we studied a terminating (“modified”) Appell function

$${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,\beta ,2\beta ;x,y)={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left(\beta \right)}_n}{{\left(2\beta \right)}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}},$$

defined for $\alpha \in {\mathbb{Z}}_{\ge 1}$ and $\beta \in {\mathbb{Z}}_{\le -1}$, jointly with the associated terminating Gauss function

$${}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ 2\beta \end{array};z\right)={\displaystyle \sum _{k=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_k{\left(\beta \right)}_k}{{\left(2\beta \right)}_k}}{\displaystyle \frac{z^k}{k!}}.$$

We demonstrated the failure of the classical Pfaff-type reduction for the non-terminating Appell function ${\mathcal{F}}_1$

$${\mathcal{F}}_1(\alpha ,\beta ,\beta ,2\beta ;x,y)={(1-y)}^{-\alpha }{}_{2}F_1\left(\begin{array}{c}\alpha ,\beta \\ 2\beta \end{array};{\displaystyle \frac{x-y}{1-y}}\right),$$

and we have given the corrected identity, ${\mathcal{CORR}}^{(\alpha ,\beta )}(x,y)$, as an explicit additional term. The strongest contribution is the derivation of an explicit closed form for the correction term, first computed in low cases (notably a = 1, 2, 3) and then stated and proved, in general, by induction on a. The final formula exhibits a structured binomial/Pascal-type pattern in its coefficients and yields several corollaries, including simplified boundary cases (for example, $\alpha =-1$).

Hypergeometric functions in mathematical analysis operate as a decisive device for comprehending many areas in pure and applied mathematics [2].

Appell [3] introduced this system of partial differential equations

$$\begin{array}{c}\hfill u(1-u)r+v(1-u)s+(\gamma -(\alpha +\beta +1\left)u\right)p-\beta vq-\alpha \beta f=0,\\ \hfill v(1-v)t+u(1-v)s+(\gamma -(\alpha +{\beta }^{\prime }+1)v)q-{\beta }^{\prime }uq-\alpha {\beta }^{\prime }f=0,\end{array}$$

(1)

in which u and v are independent variables, f is the unknown function of u and v, and $p={\displaystyle \frac{\partial f}{\partial u}}$, $q={\displaystyle \frac{\partial f}{\partial v}}$, $r={\displaystyle \frac{{\partial }^{2}f}{\partial u^2}}$, $s={\displaystyle \frac{{\partial }^{2}f}{\partial u\partial v}}$, $t={\displaystyle \frac{{\partial }^{2}f}{\partial v^2}}$. Monge’s well-known notation for partial derivatives has been investigated by many writers. The hypergeometric series in two variables

$${\mathcal{F}}_1(\alpha ,\beta ,{\beta }^{\prime },\gamma ;x,y)={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{\left(\gamma \right)}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}},\mid x\mid <1,\mid y\mid <1,$$

(2)

is a solution of (1).

Appell’s series are constituted by four double series ${\mathcal{F}}_1,{\mathcal{F}}_2,{\mathcal{F}}_3,{\mathcal{F}}_4$ of two variables that were adressed by Paul Appell and that generalize Gauss’s hypergeometric series ${}_{2}F_1$ of one variable [4,5]. Appell established the set of partial differential equations, of which these functions are solutions, and found various reduction formulas and expressions of these series in terms of hypergeometric series of one variable or in terms of elementary functions, as shown on p. 1019 in [4]

$${\mathcal{F}}_1(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right).$$

(3)

For more information about all the hypergeometric functions of two variables, we refer the reader to Kimura’s book [5], starting from page 40.

In the case where $\alpha$ is a positive integer and $\beta ,{\beta }^{\prime }$ are negative integers, we define the modified hypergeometric function ${}_{2}F_1^{\ast }$ by [6]

$${}_{2}F_1^{\ast }(\alpha ,\beta ,\beta +{\beta }^{\prime };z)={\displaystyle \sum _{k=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_k{\left(\beta \right)}_k}{{(\beta +{\beta }^{\prime })}_{k}k!}}{z}^k,\alpha \in {\mathbb{Z}}_{>0},\beta ,{\beta }^{\prime }\in {\mathbb{Z}}_{<0},$$

(4)

or the modified Appell function ${\mathcal{F}}_1^{\ast }$ of two variables by

$${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n,\alpha \in {\mathbb{Z}}_{>0},\beta ,{\beta }^{\prime }\in {\mathbb{Z}}_{<0}.$$

(5)

These two definitions are well-defined as both are two terminating, respectively, series and double series since the summations are only for $m=0,..,-\beta ,,n=0,..,-{\beta }^{\prime }$, and the fact that $\beta +{\beta }^{\prime }$ is also a negative integer does not cause any harm.

The most important question that we may ask is as follows: does ${\mathcal{F}}_1^{\ast }$ fulfill the Equation (3)? In other words, does the following equation still hold true?

$${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}})?$$

(6)

We prove that (6) is no longer true and we explicitly give, in the case where $\alpha$ is a positive integer and $\beta ,{\beta }^{\prime }$ are two negative integers, the correction term ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$, such that

$${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y).$$

(7)

In fact, the main aim of this paper is to state and prove the following result for any $\alpha \in {\mathbb{Z}}_{>0}$

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

(8)

Remark 1.

As we deal with terminating sums, we only impose $y\ne 1$. The domain of validity of the corrected identity is reduced to singular loci such as $y=1$, with continuity extension when $x=1$; since the sums are terminating, convergence is not an issue.

In the sequel, we will use the following equality:

$${}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)={\displaystyle \frac{{(\beta +{\beta }^{\prime }-\alpha )}_{\alpha }}{{({\beta }^{\prime }-\alpha )}_{\alpha }}}.$$

(9)

Proposition 1.

For any positive integer α and any negative integers β and ${\beta }^{\prime }$, we have

$${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}\ne {\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right).$$

(10)

Proof. 

The left hand side of (6),

$${\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}},$$

is a polynomial on x and y. The right hand side of (6) is given by ${\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{1-y}\right)}$. Let us denote by ${\displaystyle A_k=\frac{{\left(\alpha \right)}_k{\left(\beta \right)}_k}{k!{(\beta +{\beta }^{\prime })}_k}}$ and by $\beta =-n,n\in \mathbb{N}$; then,

$${\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{1-y}\right)=\frac{1}{{(1-y)}^{\alpha }}\left(1+A_1\frac{x-y}{1-y}+\dots +A_n{\left(\frac{x-y}{1-y}\right)}^n\right),}$$

then,

• when $x=1$, using the continuity extension, we get

  $${\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)=\frac{1}{{(1-y)}^{\alpha }}\left(1+A_1+\dots +A_n\right),}$$

  which is a rational function (which is not a polynomial),
• when $x\ne 1$, we get

  $$\begin{array}{c}{\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)=\frac{1}{{(1-y)}^{\alpha }}\left(1+A_1\frac{x-y}{1-y}+\dots +A_n{\left(\frac{x-y}{1-y}\right)}^n\right)}\hfill \\ {\displaystyle =\frac{1}{{(1-y)}^{\alpha +n}}\left({(1-y)}^n+A_1{(1-y)}^{n-1}(x-y)+\dots +A_n{(x-y)}^n\right),}\hfill \end{array}$$

  which is also a rational function and cannot be a polynomial. In practice, the degree of y in the numerator is less than or equal to n, while the degree of y in the denominator is $\alpha +n>n,\alpha \ge 1$.

□

Therefore, we have to give the correction term ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$:

$${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}})+{\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y).$$

(11)

This paper will be organized as follows. First, we show how we found our first main result ${\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)$. This result plays an important role in our second main result ${\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)$. Then, with this later result, we give our third main result ${\mathcal{CORR}}^{(3,\beta ,{\beta }^{\prime })}(x,y)$. These three steps will lead us to deduce, by induction, our final main result ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$ for any $\alpha \in \mathbb{N}$. The paper concludes with several remarks and corollaries, linking some double series to classical integer sequences.

2. The Correction Term ${\mathcal{CORR}}^{(\mathbf{1},\mathit{\beta },{\mathit{\beta }}^{\prime })}(\mathit{x},\mathit{y})$

For $\alpha =1$, we can write

$${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(1\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}={\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n.$$

The correction term ${\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)$ is given in the following theorem:

Theorem 1.

We have the following result

$${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y),$$

(12)

where ${\displaystyle {\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)=-\frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{x(1-y)}\right)}$, then we can write

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}})-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

Proof. 

Let us start from

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y),\hfill \end{array}$$

in such a way that we do NOT know ${\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)$, then the proof will be carried out using, first, the change in variables $x\leftarrow {\displaystyle \frac{1}{x}},y\leftarrow {\displaystyle \frac{1}{y}}$ in (12), which is written as

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{1}{x^m}}{\displaystyle \frac{1}{y^n}}\hfill \\ ={\displaystyle \frac{1}{(1-\frac{1}{y})}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{\frac{1}{x}-\frac{1}{y}}{1-\frac{1}{y}}}\right)+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}\left({\displaystyle \frac{1}{x}},{\displaystyle \frac{1}{y}}\right),\hfill \end{array}$$

(13)

and, second, the following three steps:

• The first step comes from the left hand side of (13):

  $${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{1}{x^m}}{\displaystyle \frac{1}{y^n}}=x^{\beta }{y}^{{\beta }^{\prime }}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n,$$
• The second step comes from the right hand side of (13) and is given by the following equation

  $${\displaystyle \frac{1}{(1-\frac{1}{y})}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{\frac{1}{x}-\frac{1}{y}}{1-\frac{1}{y}}}\right)=-{\displaystyle \frac{y}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),$$
• If we combine the first step and the second step, we get

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{(y-1)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)+x^{-\beta }{y}^{-{\beta }^{\prime }}{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}\left({\displaystyle \frac{1}{x}},{\displaystyle \frac{1}{y}}\right),\hfill \end{array}$$

  and

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y),\hfill \end{array}$$

  an identification gives:

  $${\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)=-{\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),$$

  provided that

  $$x^{-\beta }{y}^{-{\beta }^{\prime }}{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}\left({\displaystyle \frac{1}{x}},{\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right),$$

  which can be proven easily.

□

Remark 2.

Using the change IN variables $x\leftarrow {\displaystyle \frac{1}{x}},y\leftarrow {\displaystyle \frac{1}{y}}$ in ${\displaystyle \frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{1-y}\right)}$ and multiplying by $x^{-\beta }{y}^{-{\beta }^{\prime }}$, we find ${\displaystyle x^{-\beta }{y}^{-{\beta }^{\prime }}\frac{{(-1)}^{\alpha }{y}^{\alpha }}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{x(1-y)}\right).}$

• In the sequel, we adopt the following notation:

$$W^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)={\displaystyle \frac{{(-1)}^{\alpha }{x}^{-\beta }{y}^{\alpha -{\beta }^{\prime }}}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).$$

(14)

Corollary 1.

We have the following interesting particular cases:

$$\begin{array}{c}•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^{m+n}={\displaystyle \sum _{m=0}^{-\beta -{\beta }^{\prime }}}{x}^m={\displaystyle \frac{x^{1-\beta -{\beta }^{\prime }}-1}{x-1}},\hfill \\ •{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{y}^n={\displaystyle \frac{(1-\beta -{\beta }^{\prime })}{(1-{\beta }^{\prime })}}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{y}^n,\hfill \\ •{\displaystyle \sum _{m=0}^{-\beta }}\left({\displaystyle \genfrac{}{}{0pt}{}{m+0}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{(1-0)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right),\hfill \\ •{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_n}}{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)-{\displaystyle \frac{y^{-\beta -{\beta }^{\prime }+1}}{{(1-y)}^{1-\beta }}}{\displaystyle \frac{{\left({\beta }^{\prime }\right)}_{-{\beta }^{\prime }}}{{(\beta +{\beta }^{\prime })}_{-{\beta }^{\prime }}}}.\hfill \end{array}$$

Proof. 

• For $x=y$, Theorem (1) can be written as follows

$$\begin{array}{c}{\displaystyle {\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left(\genfrac{}{}{0pt}{}{m+n}{m}\right)\frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}{x}^{m+n}}\hfill \\ ={\displaystyle \frac{1}{(1-x)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};0\right)-{\displaystyle \frac{x^{-\beta -{\beta }^{\prime }+1}}{(1-x)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};0\right)\hfill \\ ={\displaystyle \frac{x^{1-\beta -{\beta }^{\prime }}-1}{x-1}}={\displaystyle \sum _{m=0}^{-\beta -{\beta }^{\prime }}}{x}^m.\hfill \end{array}$$

• For $x=1$, Theorem (1) can be written as follows

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{y}^n\hfill \\ ={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)-{\displaystyle \frac{y^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ ={\displaystyle \frac{1-y^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)=\left({\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{y}^n\right){}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right),\hfill \end{array}$$

  using (9) we get the result.
• For $y=0$, Theorem (1) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}\left({\displaystyle \genfrac{}{}{0pt}{}{m+0}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{(1-0)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0.\hfill \end{array}$$
• For $x=0$, Theorem (2) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{0+n}{0}}\right){\displaystyle \frac{{\left(\beta \right)}_0{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{0+n}}}{x}^0{y}^n\hfill \\ ={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{1-y}}\right)-{\displaystyle \frac{y^{1-{\beta }^{\prime }}}{1-y}}{\displaystyle \frac{{\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}{\left({\displaystyle \frac{y}{y-1}}\right)}^{-\beta }.\hfill \end{array}$$

  This is equivalent to

  $$\begin{array}{c}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_n}}{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)+{\displaystyle \frac{y^{1-\beta -{\beta }^{\prime }}}{{(y-1)}^{1-\beta }}}{\displaystyle \frac{{\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}.\hfill \end{array}$$

  and this is given in [6].

□

Corollary 2.

The previous equation can be written as follows:

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(1\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{\displaystyle \frac{x^m{y}^n}{m!n!}}\hfill \\ ={\displaystyle \frac{1}{(1-x)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,{\beta }^{\prime }-\beta \\ {\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x-1}}\right)+{\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{(y-1)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{{\left(\beta \right)}_{-\beta }{(x-y)}^{1-{\beta }^{\prime }}}{{(1+\beta -{\beta }^{\prime })}_{-\beta }{(1-y)}^{1-\beta }{(x-1)}^{1+\beta -{\beta }^{\prime }}}}.\hfill \end{array}$$

Proof. 

Using the result given in [6]:

$$\begin{array}{c}\hfill {}_{2}F_1^{\ast }\left(\begin{array}{c}1,-N\\ -N-N^{\prime }\end{array};z\right)={\displaystyle \frac{1}{1-z}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,-N^{\prime }+N\\ -N-N^{\prime }\end{array};{\displaystyle \frac{z}{1-z}}\right)+{\displaystyle \frac{{(-N)}_N{z}^{N^{\prime }+1}}{{\left(1\right)}_N{(N^{\prime }-N+1)}_N{(z-1)}^{N^{\prime }-N+1}}},\end{array}$$

with $-N^{\prime }+N<0$. If we substitute z with ${\displaystyle \frac{x-y}{(1-y)}}$ and multiply by ${\displaystyle \frac{1}{1-y}}$, Theorem (1) gives the desired result. □

For example, for $\alpha =1$, $\beta +{\beta }^{\prime }=-5=-4-1,-3-2,-2-3,-1-4$, we obtain

$$\begin{array}{c}•x^{4}y+{\displaystyle \frac{1}{5}}\left(x^4+4y{x}^3+2x^3+3x^{2}y+3x^2+2xy+4x+y\right)+1\hfill \\ =-{\displaystyle \frac{1}{5(x-1)}}\left(1\left({\displaystyle \frac{x-y}{x-1}}\right)+5\right)-{\displaystyle \frac{x^4{y}^2}{5(1-y)}}(1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^4+2{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^3+\hfill \\ 3{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^2+4\left({\displaystyle \frac{x-y}{x(1-y)}}\right)+5)+{\displaystyle \frac{1}{5}}{\displaystyle \frac{{(x-y)}^6}{{(x-1)}^2{(1-y)}^5}},\hfill \end{array}$$

$$\begin{array}{c}•x^3{y}^2+{\displaystyle \frac{1}{10}}\left(6x+3x^2+x^3+4y+6xy+6x^{2}y+4x^{3}y+y^2+3x{y}^2+6x^2{y}^2\right)+1\hfill \\ =-{\displaystyle \frac{1}{10(x-1)}}\left(1{\left({\displaystyle \frac{x-y}{x-1}}\right)}^2+4\left({\displaystyle \frac{x-y}{x-1}}\right)+10\right)-{\displaystyle \frac{x^3{y}^3}{10(1-y)}}(1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^3+\hfill \\ 3{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^2+6\left({\displaystyle \frac{x-y}{x(1-y)}}\right)+10)+{\displaystyle \frac{1}{10}}{\displaystyle \frac{{(x-y)}^6}{{(x-1)}^3{(1-y)}^4}},\hfill \end{array}$$

$$\begin{array}{c}•x^2{y}^3+{\displaystyle \frac{1}{10}}\left(6y+3y^2+y^3+4x+6xy+6x{y}^2+4x{y}^3+x^2+3x^{2}y+6x^2{y}^2\right)+1\hfill \\ =-{\displaystyle \frac{1}{10(x-1)}}\left(1{\left({\displaystyle \frac{x-y}{x-1}}\right)}^3+3{\left({\displaystyle \frac{x-y}{x-1}}\right)}^2+6\left({\displaystyle \frac{x-y}{x-1}}\right)+10\right)-{\displaystyle \frac{x^3{y}^3}{10(1-y)}}(1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^2\hfill \\ +4\left({\displaystyle \frac{x-y}{x(1-y)}}\right)+10)+{\displaystyle \frac{1}{10}}{\displaystyle \frac{{(x-y)}^6}{{(x-1)}^4{(1-y)}^3}},\hfill \end{array}$$

$$\begin{array}{c}•x{y}^4+{\displaystyle \frac{1}{5}}\left(y^4+4y^{3}x+2y^3+3y^{2}x+3y^2+2xy+4y+x\right)+1\hfill \\ =-{\displaystyle \frac{1}{5(x-1)}}\left(1{\left({\displaystyle \frac{x-y}{x-1)}}\right)}^4+2{\left({\displaystyle \frac{x-y}{x-1)}}\right)}^3+3{\left({\displaystyle \frac{x-y}{x-1)}}\right)}^2+4\left({\displaystyle \frac{x-y}{x-1)}}\right)+5\right)\hfill \\ -{\displaystyle \frac{x^2{y}^4}{5(1-y)}}\left(1\left({\displaystyle \frac{x-y}{x(1-y)}}\right)+5\right)+{\displaystyle \frac{1}{5}}{\displaystyle \frac{{(x-y)}^6}{{(x-1)}^5{(1-y)}^2}},\hfill \end{array}$$

and finally, with

$${\displaystyle \frac{1}{(1-x)}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,-5\\ -5\end{array};\frac{x-y}{x-1}\right)={\displaystyle \sum _{k=0}^5}{\left(\frac{x-y}{x-1}\right)}^k}$$

and

$${\displaystyle \frac{x^5{y}^6}{(y-1)}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,-0\\ -0-6\end{array};\frac{x-y}{x(1-y)}\right)=\frac{x^5{y}^6}{(y-1)},}$$

for $a=1$, $\beta +{\beta }^{\prime }=-5=-5-0$, we obtain

$$\begin{array}{c}•x^4+x^3+x^2+x+1=-{\displaystyle \frac{1}{(x-1)}}1-{\displaystyle \frac{x^{5}y}{(1-y)}}(1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^5+1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^4+\hfill \\ 1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^3+1{\left({\displaystyle \frac{x-y}{x(1-y)}}\right)}^2+1\left({\displaystyle \frac{x-y}{x(1-y)}}\right)+1)+{\displaystyle \frac{{(x-y)}^6}{(x-1){(1-y)}^6}},\hfill \end{array}$$

same analysis for $\beta +{\beta }^{\prime }=-5=-5-0$. These colored coefficients can be found in Pascal’s triangle as shown below:

1
1	1
1	2	1
1	3	3	1
1	4	6	4	1
1	5	10	10	5	1

Remark 3.

(1) We are going to find ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y),\alpha =2,3,$ such that

$${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y).$$

We will prove that the new expression $W^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$ given in (14) above will play an essential role in finding ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y),\alpha =2,3,4,\dots$.

• (2) Studying the problem for small values of (α) will help us to identify the pattern that will, subsequently, help to prove full generality.

3. The Correction Term ${\mathcal{CORR}}^{(\mathbf{2},\mathit{\beta },{\mathit{\beta }}^{\prime })}(\mathit{x},\mathit{y})$

The aim of this paragraph is to give an explicit expression of the correction term ${\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)$ such that the following equation holds

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(2\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y).\hfill \end{array}$$

We need a relation between the contiguous functions of ${\mathcal{F}}_1$. In fact, we have the following proposition given in [5], p 54.

Proposition 2.

We Have

$$\begin{array}{c}\alpha {\mathcal{F}}_1^{\ast }(\alpha +1,\beta ,{\beta }^{\prime },\gamma ;x,y)\hfill \\ =\alpha {\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\gamma ;x,y)+x{\displaystyle \frac{d}{dx}}{\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\gamma ;x,y)+y{\displaystyle \frac{d}{dy}}{\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\gamma ;x,y).\hfill \end{array}$$

(15)

The proof of this proposition is given in [5] as well. We use ${\mathcal{F}}_1^{\ast }$ instead of ${\mathcal{F}}_1$. To give an explicit expression of the correction term ${\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)$, we need the following steps.

• With $\alpha =1$ and $\gamma =\beta +{\beta }^{\prime }$ in (15), we obtain

  $$\begin{array}{c}{\mathcal{F}}_1^{\ast }(2,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)\hfill \\ ={\mathcal{F}}_1^{\ast }(1,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)+x{\displaystyle \frac{d}{dx}}{\mathcal{F}}_1^{\ast }(1,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)+y{\displaystyle \frac{d}{dy}}{\mathcal{F}}_1^{\ast }(1,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y).\hfill \end{array}$$
• Now, we use the result of our first main theorem (1):

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ +{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y);{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)=-{\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

Remark 4.

Let us denote by ${\displaystyle {\left(2F1\right)}^{(\alpha ,\beta ,{\beta }^{\prime })}=\frac{1}{{(1-y)}^{\alpha }}{}_{2}F_1^{\ast }(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{x-y}{1-y});}$ then,

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{m}}\right){\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n={\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y).\hfill \end{array}$$

• We combine these previous three steps to obtain

  $$\begin{array}{c}{\mathcal{F}}_1(2,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}(x,y)+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)\hfill \\ +x{\displaystyle \frac{d}{dx}}\left({\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y)\right)+y{\displaystyle \frac{d}{dy}}({\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}\hfill \end{array}$$

  $$\begin{array}{c}+{\mathcal{CORR}}^{(1,\beta ,{\beta }^{\prime })}(x,y))={\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}(x,y)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +x{\displaystyle \frac{d}{dx}}\left({\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}(x,y)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\right)\hfill \\ +y{\displaystyle \frac{d}{dy}}\left({\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}(x,y)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\right)\hfill \\ ={\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}(x,y)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

  $$\begin{array}{c}+x[{\displaystyle \frac{\beta }{(\beta +{\beta }^{\prime }){(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{\beta x^{-\beta -1}{y}^{1-{\beta }^{\prime }}}{(1-y)}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{\beta x^{-\beta -2}{y}^{2-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)]\hfill \\ +y[{\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{\beta (x-1)}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ -{\displaystyle \frac{(1-{\beta }^{\prime })x^{-\beta }{y}^{-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{\beta (x-1)x^{-\beta -1}{y}^{1-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)].\hfill \end{array}$$

  It is easy to prove that the terms with ${\displaystyle \frac{x-y}{1-y}}$ give

  $$\begin{array}{c}{\left(2F1\right)}^{(1,\beta ,{\beta }^{\prime })}+{\displaystyle \frac{\beta x}{(\beta +{\beta }^{\prime }){(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ +{\displaystyle \frac{y}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{\beta y(x-1)}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)={\left(2F1\right)}^{(2,\beta ,{\beta }^{\prime })},\hfill \end{array}$$

  we prove that the remaining terms with ${\displaystyle \frac{x-y}{x(1-y)}}$ give

  $$\begin{array}{c}-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)+x[{\displaystyle \frac{\beta x^{-\beta -1}{y}^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{\beta x^{-\beta -2}{y}^{2-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)]\hfill \\ +y[-{\displaystyle \frac{(1-{\beta }^{\prime })x^{-\beta }{y}^{-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{\beta (x-1)x^{-\beta -1}{y}^{1-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)]\hfill \\ =-W^{2,\beta ,{\beta }^{\prime }}(x,y)-{(\beta +{\beta }^{\prime }-2)}_1{W}^{1,\beta ,{\beta }^{\prime }}(x,y)={\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y).\hfill \end{array}$$

We can state our second main result for $\alpha =2$:

Theorem 2.

We have the following result

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y),\hfill \\ where{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)=-W^{2,\beta ,{\beta }^{\prime }}(x,y)-{(\beta +{\beta }^{\prime }-2)}_1{W}^{1,\beta ,{\beta }^{\prime }}(x,y)\hfill \\ =-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(y-1)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{(2-\beta -{\beta }^{\prime })x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

Equivalently,

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2-\beta -{\beta }^{\prime }}{0}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2-\beta -{\beta }^{\prime }}{1}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),\hfill \end{array}$$

Corollary 3.

We have the following interesting particular cases:

$$•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(2\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{x}^{m+n}={\left({\displaystyle \sum _{k=0}^{-\beta -{\beta }^{\prime }+1}}{x}^k\right)}^{\prime }={\left({\displaystyle \frac{x^{2-\beta -{\beta }^{\prime }}-1}{x-1}}\right)}^{\prime },$$

$$\begin{array}{c}•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{y}^n={\displaystyle \frac{1-y^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ -(2-\beta -{\beta }^{\prime }){\displaystyle \frac{y^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ =-{\displaystyle \frac{2(1-\beta -{\beta }^{\prime })\left(({\beta }^{\prime }-1)y^{2-{\beta }^{\prime }}-({\beta }^{\prime }-2)y^{1-{\beta }^{\prime }}-1\right)}{(2-{\beta }^{\prime }){(1-y)}^2}},\hfill \\ •{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{(1-0)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right),\hfill \\ •{\displaystyle \sum _{n=0}^{-\beta }}{\displaystyle \frac{{\left(2\right)}_n{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}}{y}^n={}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)\hfill \\ -{\displaystyle \frac{y^{-\beta -{\beta }^{\prime }+2}}{{(1-y)}^{-{\beta }^{\prime }+2}}}{\displaystyle \frac{(1-\beta ){\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}+{\displaystyle \frac{(\beta +{\beta }^{\prime }-2)y^{-\beta -{\beta }^{\prime }+1}}{{(1-y)}^{-{\beta }^{\prime }+1}}}{\displaystyle \frac{{\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}.\hfill \end{array}$$

Proof. 

• For $x=y$, Theorem (2) can be written as

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(2\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{x}^{m+n}={\displaystyle \frac{1}{{(1-x)}^2}}-{\displaystyle \frac{x^{2-\beta -{\beta }^{\prime }}}{{(x-1)}^2}}+(\beta +{\beta }^{\prime }-2){\displaystyle \frac{x^{1-\beta -{\beta }^{\prime }}}{(1-x)}}\hfill \\ ={\displaystyle \frac{1-x^{2-\beta -{\beta }^{\prime }}+(1-x)(\beta +{\beta }^{\prime }-2)x^{1-\beta -{\beta }^{\prime }}}{{(1-x)}^2}}={\left({\displaystyle \sum _{k=0}^{-\beta -{\beta }^{\prime }+1}}{x}^k\right)}^{\prime }={\left({\displaystyle \frac{x^{2-\beta -{\beta }^{\prime }}-1}{x-1}}\right)}^{\prime }.\hfill \end{array}$$

• For $x=1$, Theorem (2) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{y}^n\hfill \\ ={\displaystyle \frac{1-y^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)-(2-\beta -{\beta }^{\prime }){\displaystyle \frac{y^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right),\hfill \end{array}$$

  taking into account (9), we get the desired result.
• For $y=0$, Theorem (2) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{(1-0)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right).\hfill \end{array}$$
• For $x=0$, Theorem (2) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_n{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}}{y}^n={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{1-y}})-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(y-1)}^2}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{x(1-y)}}\right)-(2-\beta -{\beta }^{\prime }){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{x(1-y)}}\right)\hfill \\ ={}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)-{\displaystyle \frac{y^{-\beta -{\beta }^{\prime }+2}}{{(1-y)}^{-{\beta }^{\prime }+2}}}{\displaystyle \frac{(1-\beta ){\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}\hfill \\ +(\beta +{\beta }^{\prime }-2){\displaystyle \frac{y^{-\beta -{\beta }^{\prime }+1}}{{(1-y)}^{-{\beta }^{\prime }+1}}}{\displaystyle \frac{{\left(\beta \right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}.\hfill \end{array}$$

□

4. The Correction Term ${\mathcal{CORR}}^{(\mathbf{3},\mathit{\beta },{\mathit{\beta }}^{\prime })}(\mathit{x},\mathit{y})$

In this paragraph, we follow the same steps as in the case when $\alpha =2$ to obtain the explicit expression of the correction term ${\mathcal{CORR}}^{(3,\beta ,{\beta }^{\prime })}(x,y)$ such that the following equality holds,

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(3\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(3,\beta ,{\beta }^{\prime })}(x,y).\hfill \end{array}$$

For this matter, we follow the following steps.

• With $\alpha =2$ and $\beta ={\beta }^{\prime }$ in (15), we obtain

  $$\begin{array}{c}2{\mathcal{F}}_1^{\ast }(3,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)=2{\mathcal{F}}_1^{\ast }(2,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)\hfill \\ +x{\displaystyle \frac{d}{dx}}{\mathcal{F}}_1^{\ast }(2,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)+y{\displaystyle \frac{d}{dy}}{\mathcal{F}}_1^{\ast }(2,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y).\hfill \end{array}$$
• Now, we use the result of our second main theorem (2):

  $${\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\left(2F1\right)}^{(2,\beta ,{\beta }^{\prime })}(x,y)+{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y),$$

  where

  $$\begin{array}{c}{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)=-{\displaystyle \frac{x^{-\beta }{y}^{2-{\beta }^{\prime }}}{{(y-1)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{(2-\beta -{\beta }^{\prime })x^{-\beta }{y}^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$
• If we combine these previous three steps, we obtain

  $$\begin{array}{c}2{\mathcal{F}}_1^{\ast }(3,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)=2{\left(2F1\right)}^{(2,\beta ,{\beta }^{\prime })}(x,y)+2{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)\hfill \\ +x{\displaystyle \frac{d}{dx}}\left({\left(2F1\right)}^{(2,\beta ,{\beta }^{\prime })}+2{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y)\right)+y{\displaystyle \frac{d}{dy}}({\left(2F1\right)}^{(2,\beta ,{\beta }^{\prime })}\hfill \\ +{\mathcal{CORR}}^{(2,\beta ,{\beta }^{\prime })}(x,y))={\displaystyle \frac{2}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)\hfill \end{array}$$

  $$\begin{array}{c}-{\displaystyle \frac{2x^{-\beta }{y}^{2-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-2\left(2-\beta -{\beta }^{\prime }\right){\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{1-y}}\hfill \\ \times {}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)+x({\displaystyle \frac{2\beta }{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)\hfill \\ +{\displaystyle \frac{\beta x^{-\beta -1}\beta y^{2-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{2\beta x^{-2-\beta }{y}^{3-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}\hfill \\ \times {}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{(\beta +{\beta }^{\prime }-2)\beta x^{-1-\beta }{y}^{1-{\beta }^{\prime }}}{1-y}}{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{(2-\beta -{\beta }^{\prime })\beta x^{-2-\beta }{y}^{2-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right))+y({\displaystyle \frac{2}{{(1-y)}^3}}\hfill \\ \times {}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{2\beta (x-1)}{(\beta +{\beta }^{\prime }){(1-y)}^4}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)\hfill \\ -{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }}(1-{\beta }^{\prime })\left(2-\beta -{\beta }^{\prime }\right)}{(1-y)}}{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{(2-\beta -{\beta }^{\prime })x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{(2-\beta -{\beta }^{\prime })\beta (x-1)x^{-1-\beta }{y}^{1-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}\hfill \\ \times {}_{2}F_1\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{2(-{\beta }^{\prime }+2)x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{2x^{-\beta }{y}^{2-{\beta }^{\prime }}}{{(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{2\beta (x-1)x^{-1-\beta }{y}^{2-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^4}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)).\hfill \end{array}$$
• The terms with the variable ${\displaystyle \frac{x-y}{1-y}}$ are

  $${\displaystyle \frac{2}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)+x{\displaystyle \frac{2\beta }{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)$$

  $$+y\left({\displaystyle \frac{2}{{(1-y)}^3}}\right.{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{{\displaystyle \frac{2\beta (x-1)}{(\beta +{\beta }^{\prime }){(1-y)}^4}}}_2{F}_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{(1-y)}}\right)),$$

  which is, exactly, ${\displaystyle \frac{1}{{(1-y)}^3}{}_{2}F_1(3,\beta ;\beta +{\beta }^{\prime };\frac{x-y}{1-y})}$,
• the remaining terms with the variable ${\displaystyle \frac{x-y}{x(1-y)}}$ are

  $$\begin{array}{c}-{{\displaystyle \frac{2x^{-\beta }{y}^{2-{\beta }^{\prime }}{(1-y)}^2}{}}}_2{F}_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-2\left(2-\beta -{\beta }^{\prime }\right){\displaystyle \frac{x^{-\beta }{y}^{1-{\beta }^{\prime }}}{1-y}}\hfill \\ \times {}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)+x({\displaystyle \frac{\beta x^{-\beta -1}\beta y^{2-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{2\beta x^{-2-\beta }{y}^{3-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

  $$\begin{array}{c}-{\displaystyle \frac{(\beta +{\beta }^{\prime }-2)\beta x^{-1-\beta }{y}^{1-{\beta }^{\prime }}}{1-y}}{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{(2-\beta -{\beta }^{\prime })\beta x^{-2-\beta }{y}^{2-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right))\hfill \\ +y\left(-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }}(1-{\beta }^{\prime })\left(2-\beta -{\beta }^{\prime }\right)}{(1-y)}}\right.{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{(2-\beta -{\beta }^{\prime })x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

  $$\begin{array}{c}-{\displaystyle \frac{(2-\beta -{\beta }^{\prime })\beta (x-1)x^{-1-\beta }{y}^{1-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}2,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{2(2-{\beta }^{\prime })x^{-\beta }{y}^{1-{\beta }^{\prime }}}{{(1-y)}^2}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{2x^{-\beta }{y}^{2-{\beta }^{\prime }}}{{(1-y)}^3}}{}_{2}F_1\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{2\beta (x-1)x^{-1-\beta }{y}^{2-{\beta }^{\prime }}}{(\beta +{\beta }^{\prime }){(1-y)}^4}}{}_{2}F_1\left(\begin{array}{c}3,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right))\hfill \end{array}$$

  $$\begin{array}{c}=-\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{0}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+3}}{{(y-1)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{1}}\right)\hfill \\ \times {\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{2}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

We can state our third main result for $\alpha =3$:

Theorem 3.

We have the following result

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{0}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+3}}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{1}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{2}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

Corollary 4.

We have the following interesting particular cases:

$$•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(3\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{x}^{m+n}={\displaystyle \frac{1}{2}}{\left({\displaystyle \sum _{k=0}^{-\beta -{\beta }^{\prime }+2}}{x}^k\right)}^{\prime }={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x^{3-\beta -{\beta }^{\prime }}-1}{x-1}}\right)}^{″},$$

$$\begin{array}{c}•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{y}^n={\displaystyle \frac{1-y^{-\beta +3}}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ +{\displaystyle \frac{(\beta +{\beta }^{\prime }-3)}{1!}}{\displaystyle \frac{y^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ -{\displaystyle \frac{(\beta +{\beta }^{\prime }-3)(\beta +{\beta }^{\prime }-2)}{2!}}{\displaystyle \frac{y^{1-{\beta }^{\prime }}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \end{array}$$

$$\begin{array}{c}={\displaystyle \frac{\left(1-\beta -{\beta }^{\prime }\right)\left(3-\beta -{\beta }^{\prime }\right)}{\left(3-\beta \right)\left(2-\beta \right)}}\hfill \\ \times {\displaystyle \frac{(y^{3-{\beta }^{\prime }}\left(2-\beta \right)\left(1-\beta \right)-2y^{2-{\beta }^{\prime }}\left(3-\beta \right)\left(1-\beta \right)+y^{1-{\beta }^{\prime }}\left(3-\beta \right)\left(2-\beta \right)-2)}{{\left(-1+y\right)}^3}};\hfill \end{array}$$

$$•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{{(1-0)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right),$$

$$\begin{array}{c}{\displaystyle •{\displaystyle \sum _{n=0}^{-\beta }}\frac{{\left(3\right)}_n{\left(\beta \right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}{y}^n={}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{y}{y-1}\right)-\left(\frac{y^2}{{(1-y)}^2}\right.}\hfill \\ +{\displaystyle \frac{(3-\beta -{\beta }^{\prime })y}{(1-y)}}+{\displaystyle \frac{(3-\beta -{\beta }^{\prime })(2-\beta -{\beta }^{\prime })y}{2!(1-y)}}){\displaystyle \frac{{(-1)}^{\beta }{y}^{1-2{\beta }^{\prime }}{\left(\beta \right)}_{-\beta }}{{(1-y)}^{1-{\beta }^{\prime }}{\left(2\beta \right)}_{-\beta }}}.\hfill \end{array}$$

Proof. 

• For $x=y$, Theorem (3) can be written as

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\left(3\right)}_{m+n}{\displaystyle \frac{{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{x}^{m+n}={\displaystyle \frac{1}{{(1-x)}^2}}-{\displaystyle \frac{x^{2-\beta -{\beta }^{\prime }}}{{(x-1)}^2}}+(\beta +{\beta }^{\prime }-2){\displaystyle \frac{x^{1-\beta -{\beta }^{\prime }}}{(1-x)}}\hfill \\ ={\displaystyle \frac{1-x^{2-\beta -{\beta }^{\prime }}+(1-x)(\beta +{\beta }^{\prime }-2)x^{1-\beta -{\beta }^{\prime }}}{{(1-x)}^2}}={\left({\displaystyle \sum _{k=0}^{-\beta -{\beta }^{\prime }+1}}{x}^k\right)}^{\prime }={\left({\displaystyle \frac{x^{2-\beta -{\beta }^{\prime }}-1}{x-1}}\right)}^{\prime }.\hfill \end{array}$$

• For $x=1$, Theorem (3) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{y}^n={\displaystyle \frac{1}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)=-{\displaystyle \frac{y^{-{\beta }^{\prime }+2}}{{(y-1)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)\hfill \\ +(\beta +{\beta }^{\prime }-2){\displaystyle \frac{y^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right),\hfill \end{array}$$

  taking into account (9), we get the desired result.
• For $y=0$, Theorem (3) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{{(1-0)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right).\hfill \end{array}$$
• For $x=0$, Theorem (3) can be written as

  $$\begin{array}{c}{\displaystyle {\displaystyle \sum _{n=0}^{-\beta }}\frac{{\left(3\right)}_n{\left(\beta \right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}{y}^n=\frac{1}{{(1-y)}^3}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};\frac{0-y}{1-y}\right)-\frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(y-1)}^2}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{x(1-y)}}\right)-{\displaystyle \frac{(2-\beta -{\beta }^{\prime })x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{0-y}{x(1-y)}}\right)\hfill \\ ={}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)-\left({\displaystyle \frac{y^2}{{(1-y)}^2}}+{\displaystyle \frac{(3-\beta -{\beta }^{\prime })y}{(1-y)}}+{\displaystyle \frac{(3-\beta -{\beta }^{\prime })(2-\beta -\beta )y}{2!(1-y)}}\right)\hfill \\ \times {\displaystyle \frac{{(-1)}^{\beta }{y}^{1-2{\beta }^{″}}{\left(\beta \right)}_{-\beta }}{{(1-y)}^{1-{\beta }^{\prime }}{\left(2\beta \right)}_{-\beta }}}.\hfill \end{array}$$

□

5. The Correction Term ${\mathcal{CORR}}^{(\mathit{\alpha },\mathit{\beta },{\mathit{\beta }}^{\prime })}(\mathit{x},\mathit{y}),\mathit{\alpha }\in {\mathbb{Z}}_{>\mathbf{0}}$

Let us summarize. In paragraph 2, paragraph 3, and paragraph 4, we, respectively, found the following:

• for $\alpha =1$, we got

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(1\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),\hfill \end{array}$$

  (16)
• for $\alpha =2$, we got

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(2\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -\left({\displaystyle \genfrac{}{}{0pt}{}{2-\beta -{\beta }^{\prime }}{1}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),\hfill \end{array}$$
• for $\alpha =3$, we got

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(3\right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)-{\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+3}}{{(1-y)}^3}}{}_{2}F_1^{\ast }\left(\begin{array}{c}3,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{1}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+2}}{{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}2,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -\left({\displaystyle \genfrac{}{}{0pt}{}{3-\beta -{\beta }^{\prime }}{2}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+1}}{(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

• Then, by induction, we can deduce the following theorem:

Theorem 4.

For any $\alpha \in {\mathbb{Z}}_{>0}$, we have the following result

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

The change in the notations $(n,m,\beta ,{\beta }^{\prime })$ to $(i,j,-n,-m)$, respectively, gives the following

$$\begin{array}{c}{\displaystyle \sum _{j=0}^n}{\displaystyle \sum _{i=0}^m}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{(-n)}_j{(-m)}_i}{{(-n-m)}_{i+j}i!j!}}{x}^j{y}^i={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,-n\\ -n-m\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha +n+m}{\alpha -k}\right)x^n{y}^{m+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,-n\\ -n-m^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

Please see the Appendix A below.

Proof. 

The proof will be done by induction on $\alpha \ge 1$.

-

The case $\alpha =1$ is done in paragraph 1.

-

We suppose that (17) is true for any $k,1\le k\le \alpha$, and we prove it for $\alpha +1$.

-

We use (2), we take $\gamma =\beta +{\beta }^{\prime }$:

$$\begin{array}{c}\alpha {\mathcal{F}}_1^{\ast }(\alpha +1,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)\hfill \\ =\alpha {\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)+x{\displaystyle \frac{d}{dx}}{\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)+y{\displaystyle \frac{d}{dy}}{\mathcal{F}}_1^{\ast }(\alpha ,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y).\hfill \end{array}$$

-

These two later facts give

$$\begin{array}{cc}\alpha {\mathcal{F}}_1^{\ast }(\alpha +1,& \beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \frac{\alpha }{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left({}_{\mathrm{}}_{\mathrm{}}^{\mathrm{}}\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ & -\alpha {\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ & +x{\displaystyle \frac{d}{dx}}({\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ & -{\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right))\hfill \\ & +y{\displaystyle \frac{d}{dy}}({\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}})\hfill \\ & -{\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)),\hfill \end{array}$$

equivalently

$$\begin{array}{c}\alpha {\mathcal{F}}_1^{\ast }(\alpha +1,\beta ,{\beta }^{\prime },\beta +{\beta }^{\prime };x,y)={\displaystyle \frac{\alpha }{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ -\alpha {\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +x({\displaystyle \frac{\alpha \beta }{(\beta +{\beta }^{\prime }){(1-y)}^{\alpha +1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha +1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \end{array}$$

$$\begin{array}{c}+{\displaystyle \sum _{k=1}^{\alpha }}[{\displaystyle \frac{\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

$$\begin{array}{c}-{\displaystyle \frac{k\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -2}{y}^{-{\beta }^{\prime }+k+1}}{(\beta +{\beta }^{\prime }){(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\left]\right)\hfill \end{array}$$

$$\begin{array}{c}+y({\displaystyle \frac{\alpha }{{(1-y)}^{\alpha +1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{\alpha \beta (x-1)}{(\beta +{\beta }^{\prime }){(1-y)}^{\alpha +2}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha +1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{1-y}}\right)\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}[{\displaystyle \frac{(k-{\beta }^{\prime })\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k-1}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k\beta (x-1)}{(\beta +{\beta }^{\prime })}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^{k+2}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\left]\right).\hfill \end{array}$$

The terms, which are outside the three summations, give:

$${\displaystyle \frac{\alpha }{{(1-y)}^{\alpha +1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\displaystyle \frac{\alpha (x-y)}{2{(1-y)}^{\alpha +2}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha +1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right),$$

this is, exactly,

$${\displaystyle \frac{\alpha }{{(1-y)}^{\alpha +1}}}{}_{2}F_1^{\ast }(\begin{array}{c}\alpha +1,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}).$$

The terms, inside the three summations, are given by:

$$\begin{array}{c}-\alpha {\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +x\left({\displaystyle \sum _{k=1}^{\alpha }}\right[{\displaystyle \frac{\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{k\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -2}{y}^{-{\beta }^{\prime }+k+1}}{{(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\left]\right)\hfill \\ +y(-{\displaystyle \sum _{k=1}^{\alpha }}[{\displaystyle \frac{(k-{\beta }^{\prime })\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k-1}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k\beta (x-1)}{(\beta +{\beta }^{\prime })}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^{k+2}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\left]\right).\hfill \end{array}$$

Equal to

$$\begin{array}{c}-\alpha {\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

$$\begin{array}{c}+{\displaystyle \sum _{k=1}^{\alpha }}[{\displaystyle \frac{\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \frac{k\beta \left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k+1}}{{(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)]\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}[{\displaystyle \frac{(k-{\beta }^{\prime })\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \end{array}$$

$$\begin{array}{c}+{\displaystyle \frac{k\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k+1}}{{(1-y)}^{k+1}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k\beta (x-1)}{(\beta +{\beta }^{\prime })}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta -1}{y}^{-{\beta }^{\prime }+k+1}}{{(1-y)}^{k+2}}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)].\hfill \end{array}$$

Equal to

$$\begin{array}{c}{\displaystyle {\displaystyle \sum _{k=1}^{\alpha }}\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)\frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}[-\alpha {}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +\beta \end{array};\frac{x-y}{x(1-y)}\right)}\hfill \\ -\beta {}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ {}^{\prime }\beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{ky}{2(1-y)}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -(k-\beta ){}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)-{\displaystyle \frac{k}{1-y}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ +{\displaystyle \frac{k(x-1)}{2x{(1-y)}^2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)]\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)x^{-\beta }{y}^{-{\beta }^{\prime }+k}\left((\alpha -\beta -{\beta }^{\prime })(1-y)+k\right)}{{(1-y)}^{k+1}}}(k+1-y){}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right)\hfill \\ -{\displaystyle \sum _{k=1}^{\alpha }}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}}\right){\displaystyle \frac{x^{-\beta -1}{y}^{-{\beta }^{\prime }+k+1}}{{(1-y)}^{k+2}}}{\displaystyle \frac{(x-y)}{2}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k+1,\beta +1\\ \beta +{\beta }^{\prime }+1\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).\hfill \end{array}$$

This leads to

$$-\alpha {\displaystyle \sum _{k=1}^{\alpha +1}}\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha +1-\beta -{\beta }^{\prime }}{\alpha +1-k}}\right){\displaystyle \frac{x^{-\beta }{y}^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{x(1-y)}}\right).$$

□

Corollary 5.

We have the following interesting particular cases:

$$•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{x}^{m+n}={\displaystyle \frac{1}{(\alpha -1)!}}{\left({\displaystyle \frac{x^{\alpha -\beta -{\beta }^{\prime }}-1}{x-1}}\right)}^{(\alpha -1)},{(\alpha -1)}^{th}derivative,$$

$$\begin{array}{c}•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{\displaystyle \frac{{(\beta +{\beta }^{\prime }-\alpha )}_{\alpha }}{{({\beta }^{\prime }-\alpha )}_{\alpha }}}-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}{\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{\displaystyle \frac{{(\beta +{\beta }^{\prime }-k)}_k}{{({\beta }^{\prime }-k)}_k}}.\hfill \end{array}$$

$$•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{{(1-0)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right),$$

$$\begin{array}{cc}{\displaystyle •{\displaystyle \sum _{n=0}^{-\beta }}\frac{{\left(\alpha \right)}_n{\left(\beta \right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}{y}^n}\hfill & ={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-\beta -{\beta }^{\prime }+k}}{{(1-y)}^{k-\beta }}}{\displaystyle \frac{{\left(k\right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}.\hfill \end{array}$$

Proof. 

• We use Theorem (4) with $x=y$, and we proceed by induction.

• For $x=1$, the above Theorem and (9) give

  $$\begin{array}{c}•{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \sum _{n=0}^{-{\beta }^{\prime }}}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}m!n!}}{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-{\beta }^{\prime }+k}}{{(1-y)}^k}}\hfill \\ \times {}_{2}F_1^{\ast }\left(\begin{array}{c}k,\beta \\ \beta +{\beta }^{\prime }\end{array};1\right)={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{\displaystyle \frac{{(\beta +{\beta }^{\prime }-\alpha )}_{\alpha }}{{({\beta }^{\prime }-\alpha )}_{\alpha }}}-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-{\beta }^{\prime }+k}}{{(1-y)}^{k+\alpha }}}{\displaystyle \frac{{(\beta +{\beta }^{\prime }-k)}_k}{{({\beta }^{\prime }-k)}_k}}.\hfill \end{array}$$
• For $y=0$, Theorem (4) can be written as

  $$\begin{array}{c}{\displaystyle \sum _{m=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_{m+n}{\left(\beta \right)}_m{\left({\beta }^{\prime }\right)}_0}{{(\beta +{\beta }^{\prime })}_{m+0}}}{x}^m{y}^0={\displaystyle \frac{1}{{(1-0)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-0}{1-0}}\right)+0\left(True\right).\hfill \end{array}$$
• For $x=0$, Theorem (4) can be written as

  $${\displaystyle \sum _{n=0}^{-\beta }}{\displaystyle \frac{{\left(\alpha \right)}_n{\left(\beta \right)}_n}{{(\beta +{\beta }^{\prime })}_{n}n!}}{y}^n={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_{2}F_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{y}{y-1}}\right)-{\displaystyle \sum _{k=1}^{\alpha }}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\alpha -\beta -{\beta }^{\prime }}{\alpha -k}\right)y^{-\beta -{\beta }^{\prime }+k}}{{(1-y)}^{k-\beta }}}{\displaystyle \frac{{\left(k\right)}_{-\beta }}{{(\beta +{\beta }^{\prime })}_{-\beta }}}.$$

□

6. Conclusions and Open Problems

In the case where $\alpha$ is a positive integer and $\beta ,{\beta }^{\prime }$ are negative integers, we found an explicit expression of the correction term ${\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y)$, such that the following result holds true,

$$\begin{array}{c}{\displaystyle \sum _{m=0}^{-{\beta }^{\prime }}}{\displaystyle \sum _{n=0}^{-\beta }}{\left(\alpha \right)}_{m+n}{\displaystyle \frac{{\left({\beta }^{\prime }\right)}_m{\left(\beta \right)}_n}{{(\beta +{\beta }^{\prime })}_{m+n}n!m!}}{x}^m{y}^n\hfill \\ ={\displaystyle \frac{1}{{(1-y)}^{\alpha }}}{}_2\mathcal{F}_1^{\ast }\left(\begin{array}{c}\alpha ,\beta \\ \beta +{\beta }^{\prime }\end{array};{\displaystyle \frac{x-y}{1-y}}\right)+{\mathcal{CORR}}^{(\alpha ,\beta ,{\beta }^{\prime })}(x,y).\hfill \end{array}$$

(17)

The question is what is the correction term when $\alpha \in \mathbb{Q}\setminus \mathbb{Z}$?

I considered the formula in [4], pp. 1018–1019. The very next section 9.183 starts with six equal F1-functions, and generally there are ten such groups of six F1-solutions (similarly to 4 × 6 Kummer’s 2F1 solutions to the classical hypergeometric equation). One can examine the generation of “corrections” for the 6 × 10 structure of F1-solutions similarly as in [6].