Combinatorial Identities Concerning Binomial Quotients

Abstract

Making use of a telescoping approach, three types of sums of binomial quotients are examined. The summation terms of the two types of alternating sums have symmetry (i.e., their numerators and denominators are completely symmetric). We obtained a series of their explicit sums. Furthermore, by means of binomial relations, three recurrence relations of the sums are derived. In addition, series of double summation formulae involving binomial quotients are established.

1. Introduction and Motivation

Combinatorial identities have widespread applications in many fields, such as mathematics, computer sciences and physics. In [1], Gould recorded more than 500 binomial identities. Among many others, he listed some interesting summation formulae of binomial quotients, such as [1] Equations (4.22), (4.26), (4.23) and (4.27):

$$\begin{array}{cc}\hfill & \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}{\left(\genfrac{}{}{0pt}{}{2m}{2k}\right)}=\frac{1+{(-1)}^m}{2}·\frac{2m+1}{m+1};\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill & \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m}{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{1+{(-1)}^m}{2}·\frac{1}{1-m};\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}{\left(\genfrac{}{}{0pt}{}{2m+1}{2k+1}\right)}=\frac{1-{(-1)}^m}{2}·\frac{1}{m+2}+{(-1)}^m;\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill & \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m+1}{2k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{1-{(-1)}^m}{2m}+1.\hfill \end{array}$$

(4)

For more combinatorial identities with inverse binomial coefficients, the reader can refer to [2,3,4,5,6,7,8]. In particular, the following results are well known:

$$\begin{array}{c}\hfill {\mathcal{S}}_m:=\sum _{k=0}^m{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}^{-1}=\frac{m+1}{2^m}\sum _{k=0}^m\frac{2^k}{k+1}.\end{array}$$

(5)

Furthermore, ${\mathcal{S}}_m$ satisfies the recurrence relation

$$\begin{array}{c}\hfill {\mathcal{S}}_m=1+\frac{m+1}{2m}{\mathcal{S}}_{m-1}.\end{array}$$

(6)

The above results are the subject of problem 1 in the afternoon session of the 1958 Putnam Exam. They were recorded by Comtet [9] (Exercise 15, p. 294) and Graham et al. [10] (Exercise 5.100, pp. 542–543). There are different proofs for the results, which can be found in [2,3,4,5]. In addition, Mansour [2] also obtained the following weighted result:

$$\sum _{k=0}^m\frac{k}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{(m+1)(2^m-1)}{2^m}+\frac{1}{2^{m+1}}\sum _{k=0}^{m-2}\frac{(m-k)(m-k-1)2^k}{k+1}.$$

Among many methods, such as the generating function method [11,12,13], WZ method [14,15], Riordan array approach [16,17] and hypergeometric series technique [18], to prove combinatorial identities, the telescoping technique proposed by Zeilberger [19,20] is a quite efficient approach to computing binomial sums. This is the main method used in this paper, and thus we will give a brief introduction to it here.

For summing ${\displaystyle \sum _{k=\ell }^m}{a}_k$, we construct a new sequence $A_k$ such that $a_k=A_{k+1}-A_k$, and then

$$\begin{array}{c}\hfill \sum _{k=\ell }^m{a}_k=\sum _{k=1}^m(A_{k+1}-A_k)=A_{m+1}-A_{\ell }.\end{array}$$

Recently, Chu and Guo [21] examined, by means of telescoping [22] and the linearization method [23], the following alternating sums of binomial quotients:

$$\begin{array}{c}\hfill {\mathrm{Q}}_{\lambda ,\epsilon }\left(m\right):=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)},\end{array}$$

(7)

where $\lambda ,m\in {\mathbb{N}}_0$ and $\epsilon \in \{0,1\}$.

Motivated by Equations (1)–(4) and (7) above, in this paper, we shall, by means of telescoping, calculate the following sums of binomial quotients:

$${\Omega }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k}\right)},\mathrm{with}\tau ,\epsilon \in {\mathbb{N}}_0.$$

$$\begin{array}{c}\hfill {\Phi }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)},\mathrm{with}\delta \ge \epsilon .\end{array}$$

$${\Psi }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)},\mathrm{with}\delta \le \epsilon .$$

As we can see, the summation terms of ${\Phi }_m(\tau ,\delta ,\epsilon )$ and ${\Psi }_m(\tau ,\delta ,\epsilon )$ have symmetry (i.e., their numerators and denominators are completely symmetrical).

Furthermore, we shall further explore the following double finite series:

$$\begin{array}{c}\hfill {\mathrm{P}}_{\lambda ,\epsilon }\left(m\right):=\sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)},\mathrm{with}\lambda \le m,\epsilon =0,1,\end{array}$$

whose results are related to ${\mathrm{Q}}_{\lambda ,\epsilon }\left(m\right)$.

For an indeterminate x, the symbol ${\left(x\right)}_n$ will be used in the next sections, with the rising factorial defined by quotients of the gamma function

$${\left(x\right)}_n=\frac{\Gamma (x+n)}{\Gamma \left(x\right)}=\left\{\begin{array}{cc}1,\hfill & n=0,\hfill \\ x(x+1)(x+2)\cdots (x+n-1),\hfill & n=1,2,\cdots .\hfill \end{array}\right.$$

The generalized hypergeometric series used by Bailey [24] are defined as follows:

$$\begin{array}{c}\hfill {}_{1+p}F_q\left[\begin{array}{c}{a}_0,a_1,\cdots ,a_p\\ b_1,\cdots ,b_q\end{array}|z\right]=\sum _{k=0}^{\infty }\frac{{\left(a_0\right)}_k{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k}{k!{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k}.\end{array}$$

Then, the above three sums can be rewritten as hypergeometric series ${}_{3}F_2$:

$$\begin{array}{cc}& {\Omega }_m(\tau ,\delta ,\epsilon )=\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m}\right){}_{3}F_2\left[\begin{array}{ccc}-m,& 1,& 1\\ & -m-\epsilon ,& (\tau -1)m+\delta +1\end{array}|1\right],\hfill \\ & {\Phi }_m(\tau ,\delta ,\epsilon )=\frac{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta }{\epsilon }\right)}{}_{3}F_2\left[\begin{array}{ccc}-m,& \epsilon +1,& 1\\ & -\tau m,& \epsilon -\delta -m\end{array}|1\right],\hfill \\ & {\Psi }_m(\tau ,\delta ,\epsilon )=\frac{\left(\genfrac{}{}{0pt}{}{m+\delta }{\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}{}_{3}F_2\left[\begin{array}{ccc}-\tau m,& \epsilon -\delta -m,& 1\\ & -m,& \epsilon +1\end{array}|1\right].\hfill \end{array}$$

In particular, when setting $\tau =1,\delta -\epsilon =2$, the sum ${\Omega }_m(1,\delta ,\epsilon )$ can be evaluated with Saalschütz’s theorem. Furthermore, these series are difficult to calculate using the classical Saalschütz, Dixon, Watson and Whipple theorems. Many scholars have studied the hypergeometric series concerning ${}_{3}F_2$, such as Chu [23], Chen [25], Li and Chu [26] and Milgram [27]. However, the series examined in this paper are also difficult to calculate directly from their results.

2. Summation Formulae ${\Omega }_{\mathit{m}}(\tau ,\delta ,\epsilon )$

In this section, we shall evaluate ${\Omega }_m(\tau ,\delta ,\epsilon )$, defined by the binomial quotient

$${\Omega }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k}\right)},\mathrm{with}\tau ,\epsilon \in {\mathbb{N}}_0.$$

Theorem 1. 

${\Omega }_m(\tau ,\delta ,0)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{(\tau -1)m+\delta }{(\tau -1)m+\delta -1}\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m}\right)-\frac{m+1}{(\tau -1)m+\delta -1}.\end{array}$$

(8)

Proof. 

Notice that the quotient can be rewritten as

$$\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{\tau m+\delta }{\tau m-m+\delta -1}({\mathcal{A}}_{k+1}-{\mathcal{A}}_k)-\frac{\tau m+\delta }{\tau m-m+\delta -1}({\mathrm{A}}_{k+1}-{\mathrm{A}}_k),$$

where

$${\mathcal{A}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta -1}{m-k-1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}\mathrm{and}{\mathrm{A}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}.$$

By means of the telescoping technique, the sum becomes

$$\begin{array}{cc}\hfill \sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}& =1+\frac{\tau m+\delta }{m}+\sum _{k=0}^{m-2}\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}\hfill \\ & =1+\frac{\tau m+\delta }{m}+\frac{\tau m+\delta }{\tau m-m+\delta -1}({\mathcal{A}}_{m-1}-{\mathcal{A}}_0-{\mathrm{A}}_{m-1}+{\mathrm{A}}_0).\hfill \end{array}$$

The proof follows from the fact that

$${\mathcal{A}}_{m-1}=\frac{1}{m},{\mathcal{A}}_0=\left(\genfrac{}{}{0pt}{}{\tau m+\delta -1}{m-1}\right)\mathrm{and}{\mathrm{A}}_{m-1}=\frac{\tau m+\delta }{m},{\mathrm{A}}_0=\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m}\right).$$

□

Specially, we have the following particular cases from Theorem 1.

Corollary 1. 

$$\begin{array}{cc}\hfill {\Omega }_m(1,0,0)& =\sum _{k=0}^{m}1=m+1;\hfill \\ \hfill {\Omega }_m(2,1,0)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{2m+1}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}=\frac{m+1}{m}\left\{\left(\genfrac{}{}{0pt}{}{2m+1}{m}\right)-1\right\}.\hfill \end{array}$$

Theorem 2. 

${\Omega }_m(\tau ,\delta ,1)$:

$$\begin{array}{cc}\hfill \sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}& =\frac{m+2}{{(m(\tau -1)+\delta -2)}_2}\hfill \\ & +\frac{(\tau -1)m^2+m(\tau +\delta -3)+\delta -3}{{(m(\tau -1)+\delta -2)}_2}\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right).\hfill \end{array}$$

(9)

Proof. 

Write the summand as

$$\begin{array}{cc}\hfill \frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}& =\frac{m+2}{(\tau -1)m+\delta -2}({\mathcal{B}}_{k+1}-{\mathcal{B}}_k)\hfill \\ & -\frac{(\tau -1)m^2+m(\tau +2\delta -4)+2\delta -4}{{(m(\tau -1)+\delta -2)}_2}({\mathrm{B}}_{k+1}-{\mathrm{B}}_k),\hfill \end{array}$$

where

$${\mathcal{B}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}\mathrm{and}{\mathrm{B}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}.$$

By using telescoping, we can evaluate the sum

$$\begin{array}{cc}\hfill \sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}& =\frac{m+2}{(\tau -1)m+\delta -2}({\mathcal{B}}_{m+1}-{\mathcal{B}}_0)\hfill \\ & -\frac{(\tau -1)m^2+m(\tau +2\delta -4)+2\delta -4}{{(m(\tau -1)+\delta -2)}_2}({\mathrm{B}}_{m+1}-{\mathrm{B}}_0).\hfill \end{array}$$

Then, the proof follows from the fact that ${\mathcal{B}}_{m+1}={\mathrm{B}}_{m+1}=1$ and

$${\mathcal{B}}_0=\frac{1}{m+2}\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right)\mathrm{and}{\mathrm{B}}_0=\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right).$$

□

For spacial cases, we obtain from Theorem 2 the corollary below.

Corollary 2. 

$$\begin{array}{cc}\hfill {\Omega }_m(2,3,1)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{2m+3}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{1}{m+1}+\frac{m}{m+1}\left(\genfrac{}{}{0pt}{}{2m+3}{m+1}\right);\hfill \\ \hfill {\Omega }_m(3,3,1)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{3m+3}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{m+2}{{(2m+1)}_2}+\frac{2m^2+3m}{{(2m+1)}_2}\left(\genfrac{}{}{0pt}{}{3m+3}{m+1}\right);\hfill \\ \hfill {\Omega }_m(4,3,1)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{4m+3}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{m+2}{{(3m+1)}_2}+\frac{3m^2+4m}{{(3m+1)}_2}\left(\genfrac{}{}{0pt}{}{4m+3}{m+1}\right).\hfill \end{array}$$

Theorem 3. 

The recurrence relation is expressed as

$$\begin{array}{c}\hfill {\Omega }_m(\tau ,\delta ,\epsilon )=\frac{m+\epsilon +1}{m+\epsilon }{\Omega }_m(\tau ,\delta ,\epsilon -1)-{\Omega }_{m+1}(\tau ,\delta -\tau ,\epsilon -1)+\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right).\end{array}$$

(10)

Proof. 

By means of the relation

$${\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k}\right)}^{-1}={\left(\genfrac{}{}{0pt}{}{m+\epsilon -1}{k}\right)}^{-1}-\frac{k}{m+\epsilon }{\left(\genfrac{}{}{0pt}{}{m+\epsilon -1}{k}\right)}^{-1},$$

the sum can then be rewritten as

$$\begin{array}{c}\hfill \sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k}\right)}=\frac{m+\epsilon +1}{m+\epsilon }\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon -1}{k}\right)}-\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k+1}\right)}.\end{array}$$

By replacing k with $k-1$, the second sum on the right-hand side becomes

$${\mathrm{V}}_{m+1}:=\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k+1}\right)}=\sum _{k=1}^{m+1}\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon }{k}\right)}.$$

Thus, we have

$${\mathrm{V}}_m=\sum _{k=1}^m\frac{\left(\genfrac{}{}{0pt}{}{\tau m-\tau +\delta }{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\epsilon -1}{k}\right)}={\Omega }_m(\tau ,\delta -\tau ,\epsilon -1)-\left(\genfrac{}{}{0pt}{}{\tau m-\tau +\delta }{m}\right),$$

which confirms the recurrence relation stated in Theorem 3. □

It is not difficult to find the corollary below by means of Theorems 2 and 3.

Corollary 3. 

$$\begin{array}{cc}\hfill {\Omega }_m(2,3,2)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{2m+3}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k}\right)}=\frac{m^3+2m^2-m+2}{{\left(m\right)}_3}\left(\genfrac{}{}{0pt}{}{2m+3}{m+1}\right)-\frac{2(m+3)}{{\left(m\right)}_3};\hfill \\ \hfill {\Omega }_m(2,3,3)& =\sum _{k=0}^m\frac{\left(\genfrac{}{}{0pt}{}{2m+3}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+3}{k}\right)}=\frac{m^2(m^2-1)(m+4)+12(m-2)}{{(m-1)}_4(m-1)}\left(\genfrac{}{}{0pt}{}{2m+3}{m+1}\right)+\frac{6(m+4)}{{(m-1)}_4}.\hfill \end{array}$$

3. Summation Formulae ${\Phi }_{\mathit{m}}(\tau ,\delta ,\epsilon )$

In this section, we shall deal with the sum ${\Phi }_m(\tau ,\delta ,\epsilon )$ defined by the alternating binomial quotient:

$$\begin{array}{c}\hfill {\Phi }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)},\mathrm{with}\delta \ge \epsilon .\end{array}$$

Theorem 4. 

${\Phi }_m(\tau ,1,1)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}={(-1)}^m\frac{m+2}{\tau m+3}+\frac{1}{\tau m+3}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right).\end{array}$$

(11)

Proof. 

We write the summand as

$$\begin{array}{c}\hfill \frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}=\frac{\tau m-m+1}{\tau m+3}({\mathrm{C}}_k+{\mathrm{C}}_{k+1}),\mathrm{where}{\mathrm{C}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k+1}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}.\end{array}$$

By using telescoping, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}& ={(-1)}^m+\sum _{k=0}^{m-1}{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}\hfill \\ & ={(-1)}^m+\frac{\tau m-m+1}{\tau m+3}\{{\mathrm{C}}_0-{(-1)}^m{\mathrm{C}}_m\}.\hfill \end{array}$$

Then the proof follows from the fact that

$${\mathrm{C}}_0=\frac{1}{m+1}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m}\right)\mathrm{and}{\mathrm{C}}_m=1.$$

□

Theorem 5. 

${\Phi }_m(\tau ,1,0)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{(m+1)(\tau m+3)+1}{{(\tau m+2)}_2}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)+{(-1)}^m\frac{m+2}{{(\tau m+2)}_2}.\end{array}$$

(12)

Proof. 

We write the summand as

$$\begin{array}{c}\hfill \frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{m+2}{\tau m+2}({\mathrm{D}}_k+{\mathrm{D}}_{k+1})-\frac{m+2}{\tau m+3}({\mathcal{D}}_k+{\mathcal{D}}_{k+1}),\end{array}$$

where

$$\begin{array}{c}\hfill {\mathrm{D}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k+1}{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}\mathrm{and}{\mathcal{D}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k+1}{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}.\end{array}$$

By using telescoping, we can evaluate the sum

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{m+2}{\tau m+2}\{{\mathrm{D}}_0+{(-1)}^m{\mathrm{D}}_{m+1}\}-\frac{m+2}{\tau m+3}\{{\mathcal{D}}_0+{(-1)}^m{\mathcal{D}}_{m+1}\}.\end{array}$$

Then, the proof follows from the fact that ${\mathrm{D}}_{m+1}={\mathcal{D}}_{m+1}=1$ and

$${\mathrm{D}}_0=\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)\mathrm{and}{\mathcal{D}}_0=\frac{1}{m+2}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right).$$

□

By setting $\tau =2$, we obtain from Theorems 4 and 5 the following corollary.

Corollary 4. 

$$\begin{array}{cc}\hfill {\Phi }_m(2,1,1)& =\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}={(-1)}^m\frac{m+2}{2m+3}+\frac{1}{2m+3}\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right);\hfill \\ \hfill {\Phi }_m(2,1,0)& =\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}=\frac{(m+1)(2m+3)+1}{{(2m+2)}_2}\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right)+{(-1)}^m\frac{m+2}{{(2m+2)}_2}.\hfill \end{array}$$

Theorem 6. 

${\Phi }_m(\tau ,2,2)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}=\frac{2}{(m+2)(\tau m+4)}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)+{(-1)}^m\frac{m+3}{\tau m+4}.\end{array}$$

(13)

Proof. 

By keeping in mind the summation term

$$\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}=({\mathrm{E}}_{k+1}+{\mathrm{E}}_k)-\frac{m+3}{\tau m+4}({\mathcal{E}}_{k+1}+{\mathcal{E}}_k),$$

where

$${\mathrm{E}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}\mathrm{and}{\mathcal{E}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+3}{k+3}\right)},$$

then by means of telescoping, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}& ={(-1)}^m+\sum _{k=0}^{m-1}{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}\hfill \\ & ={(-1)}^m+{\mathrm{E}}_0-{(-1)}^m{\mathrm{E}}_m-\frac{m+3}{\tau m+4}({\mathcal{E}}_0-{(-1)}^m{\mathcal{E}}_m).\hfill \end{array}$$

By substituting the values ${\mathrm{E}}_m={\mathcal{E}}_m=1$ and

$${\mathrm{E}}_0=\frac{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{2}\right)}\mathrm{and}{\mathcal{E}}_0=\frac{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}{\left(\genfrac{}{}{0pt}{}{m+3}{3}\right)}$$

into the above equation, we can then find the desired identity. □

Theorem 7. 

${\Phi }_m(\tau ,2,1)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}=\frac{\tau m^2+\tau m+4m+6}{(m+2){(\tau m+3)}_2}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)+{(-1)}^m\frac{m+3}{{(\tau m+3)}_2}.\end{array}$$

(14)

Proof. 

We rewrite the summand as

$$\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}=\frac{m+3}{\tau m+3}({\mathrm{F}}_{k+1}+{\mathrm{F}}_k)-\frac{m+3}{\tau m+4}({\mathcal{F}}_{k+1}+{\mathcal{F}}_k),$$

where

$${\mathrm{F}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k+1}{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}\mathrm{and}{\mathcal{F}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k+1}{m-k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m+3}{k+2}\right)},$$

Then, by making use of the telescoping approach, we obtain

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}=\frac{m+3}{\tau m+3}({\mathrm{F}}_0+{(-1)}^m{\mathrm{F}}_{m+1})-\frac{m+3}{\tau m+4}({\mathcal{F}}_0+{(-1)}^m{\mathcal{F}}_{m+1}).\end{array}$$

The proof follows from the fact that ${\mathrm{F}}_{m+1}={\mathcal{F}}_{m+1}=1$ and

$${\mathrm{F}}_0=\frac{1}{m+2}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)\mathrm{and}{\mathcal{F}}_0=\frac{2}{(m+2)(m+3)}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right).$$

□

For spacial cases, when $\tau =2$, we obtain from Theorems 6 and 7 the corollary below.

Corollary 5. 

$$\begin{array}{cc}\hfill {\Phi }_m(2,2,2)& =\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+2}\right)}=\frac{1}{{(m+2)}^2}\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right)+{(-1)}^m\frac{m+3}{2m+4};\hfill \\ \hfill {\Phi }_m(2,2,1)& =\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{2m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k+1}\right)}=\frac{m^2+3m+3}{{(m+2)}^2(2m+3)}\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right)+{(-1)}^m\frac{m+3}{{(2m+3)}_2}.\hfill \end{array}$$

Theorem 8. 

The recurrence relation is expressed as

$$\begin{array}{c}\hfill {\Phi }_m(\tau ,\delta ,\epsilon )=\frac{m+\delta +1}{m+\delta }{\Phi }_m(\tau ,\delta -1,\epsilon -1)-{\Phi }_m(\tau ,\delta ,\epsilon -1).\end{array}$$

(15)

Proof. 

By means of the relation

$${\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)}^{-1}={\left(\genfrac{}{}{0pt}{}{m+\delta -1}{k+\epsilon }\right)}^{-1}-\frac{k+\epsilon }{m+\delta }{\left(\genfrac{}{}{0pt}{}{m+\delta -1}{k+\epsilon }\right)}^{-1},$$

we have

$$\begin{array}{cc}\hfill {\Phi }_m(\tau ,\delta ,\epsilon )& :=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)}\hfill \\ & =\frac{m+\delta +1}{m+\delta }\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta -1}{k+\epsilon }\right)}-\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon +1}\right)}\hfill \\ & =\frac{m+\delta +1}{m+\delta }{\Phi }_m(\tau ,\delta -1,\epsilon )-{\Phi }_m(\tau ,\delta ,\epsilon +1).\hfill \end{array}$$

Then, the proof follows by making the replacement $\epsilon \to \epsilon -1$. □

By means of Theorem 8 above, we obtain from Theorem 5 and Theorem 7 the following corollary.

Corollary 6 

(${\Phi }_m(\tau ,2,0)$).

$$\begin{array}{cc}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}{\left(\genfrac{}{}{0pt}{}{m+2}{k}\right)}& =\frac{(m+1)(\tau m+4)\left[\right(m+2\left)\right(\tau m+3)+2]+4}{(m+2){(\tau m+2)}_3}\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)\hfill \\ & +{(-1)}^m\frac{2(m+3)}{{(\tau m+2)}_3}.\hfill \end{array}$$

4. Summation Formulae ${\Psi }_{\mathit{m}}(\mathit{\tau },\mathit{\delta },\mathit{\epsilon })$

In this section, we shall compute the following alternating sum:

$${\Psi }_m(\tau ,\delta ,\epsilon ):=\sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)},\mathrm{with}\delta \le \epsilon .$$

Theorem 9. 

${\Psi }_m(\tau ,1,1)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}=\frac{1}{\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)}.\end{array}$$

(16)

Proof. 

Keep in mind that the summand can be rewritten as

$$\frac{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}=\frac{m+1}{\tau m+1}({\mathcal{G}}_k+{\mathcal{G}}_{k+1}),\mathrm{where}{\mathcal{G}}_k=\frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}.$$

Using telescoping approach, the sum can be evaluated:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+1}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}=\frac{m+1}{\tau m+1}\{{\mathcal{G}}_0+{(-1)}^m{\mathcal{G}}_{m+1}\}.\end{array}$$

Then, the proof follows from the fact that ${\mathcal{G}}_{m+1}=0$ and ${\mathcal{G}}_0={\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}^{-1}$. □

Theorem 10. 

${\Psi }_m(\tau ,0,1)$:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}=\frac{1}{\left(\genfrac{}{}{0pt}{}{\tau m+1}{m+1}\right)}-{(-1)}^m\frac{\tau -1}{\tau (\tau m+1)}.\end{array}$$

(17)

Proof. 

Notice that the summand can be rewritten as

$$\frac{\left(\genfrac{}{}{0pt}{}{m}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}=\frac{m+1}{\tau m+1}({\mathrm{H}}_k+{\mathrm{H}}_{k+1})-\frac{1}{\tau }({\mathcal{H}}_k+{\mathcal{H}}_{k+1}),$$

where

$${\mathrm{H}}_k=\frac{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}\mathrm{and}{\mathcal{H}}_k=\frac{\left(\genfrac{}{}{0pt}{}{m-1}{k-1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}.$$

By means of telescoping, we obtain

$$\begin{array}{cc}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}& =\frac{m}{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}+\sum _{k=1}^{m-1}{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m}{k+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m-k}{m-k}\right)}\hfill \\ & =\frac{m}{\left(\genfrac{}{}{0pt}{}{\tau m}{m}\right)}+\frac{m+1}{\tau m+1}\{{\mathrm{H}}_1-{(-1)}^m{\mathrm{H}}_m\}-\frac{1}{\tau }\{{\mathcal{H}}_1-{(-1)}^m{\mathcal{H}}_m\}.\hfill \end{array}$$

Then, the proof can be completed with the fact that ${\mathcal{H}}_m={\mathrm{H}}_m=1$ and

$${\mathcal{H}}_1=\frac{1}{\left(\genfrac{}{}{0pt}{}{\tau m-1}{m-1}\right)}\mathrm{and}{\mathrm{H}}_1=\frac{m}{\left(\genfrac{}{}{0pt}{}{\tau m-1}{m-1}\right)}.$$

□

Theorem 11. 

The recurrence relation is expressed as

$$\begin{array}{c}\hfill {\Psi }_m(\tau ,\delta ,\epsilon )={\Psi }_m(\tau ,\delta -1,\epsilon -1)+{\Psi }_m(\tau ,\delta -1,\epsilon ).\end{array}$$

(18)

Proof. 

The proof follows by using the relation

$$\left(\genfrac{}{}{0pt}{}{m+\delta }{k+\epsilon }\right)=\left(\genfrac{}{}{0pt}{}{m+\delta -1}{k+\epsilon -1}\right)+\left(\genfrac{}{}{0pt}{}{m+\delta -1}{k+\epsilon }\right).$$

□

5. Double Summation Formulae ${\mathrm{P}}_{\lambda ,\epsilon }\left(\mathit{m}\right)$

In this section, we shall explore the following double summation formulae:

$$\begin{array}{c}\hfill {\mathrm{P}}_{\lambda ,\epsilon }\left(m\right):=\sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)},\mathrm{with}\lambda \le m,\epsilon =0,1.\end{array}$$

First, we prove the following lemma.

Lemma 1. 

For integers $m,\lambda$ and j, the sequence

$$\begin{array}{c}\hfill {\mathrm{S}}_{m,\lambda }\left(x\right):=\sum _{j=k}^m\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j}\right)}\end{array}$$

satisfies the recurrence relation

$$\begin{array}{c}\hfill (1+x){\mathrm{S}}_{m+1,\lambda }\left(x\right)=x\frac{m+\lambda +2}{m+\lambda +1}{\mathrm{S}}_{m,\lambda }\left(x\right)+\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k}\right)}+\frac{x^{m+2}}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{m+1}\right)}.\end{array}$$

(19)

Proof. 

By making use of the binomial recurrence relation

$$\frac{1}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{j}\right)}=\frac{1}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j-1}\right)}-\frac{m+\lambda +1-j}{m+\lambda +2-j}\frac{1}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{j-1}\right)},$$

we can manipulate the sum

$$\begin{array}{cc}\hfill {\mathrm{S}}_{m+1,\lambda }\left(x\right)& =\sum _{j=k}^{m+1}\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+1+\lambda }{j}\right)}=\sum _{j=k}^{m+1}{x}^j\left\{\frac{1}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j-1}\right)}-\frac{m+\lambda +1-j}{m+\lambda +2-j}\frac{1}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{j-1}\right)}\right\}\hfill \\ & =\sum _{j=k}^{m+1}\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j-1}\right)}-\sum _{j=k}^{m+1}\left(1-\frac{1}{m+\lambda +2-j}\right)\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{j-1}\right)}.\hfill \end{array}$$

In light of the binomial relation

$$(m+\lambda +2-j)\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{j-1}\right)=(m+\lambda +1)\left(\genfrac{}{}{0pt}{}{m+\lambda }{j-1}\right),$$

we obtain that

$$\begin{array}{c}\hfill {\mathrm{S}}_{m+1,\lambda }\left(x\right)=\frac{m+\lambda +2}{m+\lambda +1}\sum _{j=k}^{m+1}\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j-1}\right)}-\sum _{j=k}^{m+1}\frac{x^j}{\left(\genfrac{}{}{0pt}{}{m+1+\lambda }{j-1}\right)}.\end{array}$$

Then, by taking the replacement of $j\to j+1$, we have

$$\begin{array}{cc}\hfill {\mathrm{S}}_{m+1,\lambda }\left(x\right)& =\frac{m+\lambda +2}{m+\lambda +1}\sum _{j=k-1}^{m+1}\frac{x^{j+1}}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{j}\right)}-\sum _{j=k-1}^{m+1}\frac{x^{j+1}}{\left(\genfrac{}{}{0pt}{}{m+1+\lambda }{j}\right)}\hfill \\ & =\frac{m+\lambda +2}{m+\lambda +1}\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{k-1}\right)}+x\frac{m+\lambda +2}{m+\lambda +1}{\mathrm{S}}_{m,\lambda }\left(x\right)-\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k-1}\right)}+\frac{x^{m+2}}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{m+1}\right)}-x{\mathrm{S}}_{m+1,\lambda }\left(x\right).\hfill \end{array}$$

The proof can be completed by using

$$\begin{array}{cc}\hfill \frac{m+\lambda +2}{m+\lambda +1}\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda }{k-1}\right)}-\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k-1}\right)}& =\frac{m+\lambda +2}{k}\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k}\right)}-\frac{m+\lambda +2-k}{k}\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k}\right)}\hfill \\ & =\frac{x^k}{\left(\genfrac{}{}{0pt}{}{m+\lambda +1}{k}\right)}.\hfill \end{array}$$

□

Corollary 7. 

For two integers m and k, it holds that

$$\begin{array}{c}\hfill \sum _{j=k}^m{(-1)}^j\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}=\frac{m+1}{m+2}\left\{{(-1)}^m+\frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}\right\}.\end{array}$$

(20)

Proof. 

By letting $x=-1$ and $\lambda =0$ in the recurrence relation in Equation (19), we obtain that

$$0=\frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}+{(-1)}^m-\frac{m+2}{m+1}{\mathrm{S}}_{m,0}(-1),$$

which yields the desired result

$${\mathrm{S}}_{m,0}(-1)=\sum _{j=k}^m{(-1)}^j\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}=\frac{m+1}{m+2}\left\{{(-1)}^m+\frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)}\right\}.$$

□

Theorem 12. 

For integers $\lambda \le m$ and $\epsilon \in 0,1$, it holds that

$$\begin{array}{c}\hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}=2^{\lambda -1}{(-2)}^m\frac{m+1}{m+2}+\frac{m+1}{m+2}{\Omega }_{\lambda -1,\epsilon }(m+1).\end{array}$$

(21)

Proof. 

Exchanging the order of summation and applying Corollary 7 yields

$$\begin{array}{cc}\hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)\sum _{k=0}^j\frac{{(-1)}^j}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}\hfill \\ & ={(-1)}^m\frac{m+1}{m+2}\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)+\frac{m+1}{m+2}{\Omega }_{\lambda -1,\epsilon }(m+1).\hfill \end{array}$$

Then, the proof can be completed as follows:

$$\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m+\lambda }{2k+\epsilon }\right)=2^{m+\lambda -1},\lambda \le m,\epsilon =0,1.$$

□

By employing the results of ${\Omega }_{\lambda ,\epsilon }\left(m\right)$ in [21], we can obtain series of finite double summation formulae on ${\mathrm{P}}_{\lambda ,\epsilon }\left(m\right)$. For instance, if we let $\lambda =4,3,2,1$, then we have the following corollaries.

Corollary 8. 

For $\epsilon =0$, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+4}{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{8{(-2)}^m-1\right\}+\frac{m^2-m+4}{m+2};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+3}{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{4{(-2)}^m-1\right\}+\frac{m-1}{m+2};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+2}{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{2{(-2)}^m+1\right\}-\frac{1}{2}\sum _{k=0}^m\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+1}{2k}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{{(-2)}^m+\frac{m+3}{2}\right\}-\frac{m+1}{4}\sum _{k=0}^m\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}.\hfill \end{array}$$

Corollary 9. 

For $\epsilon =1$, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+4}{2k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{8{(-2)}^m+2\right\}-\frac{2}{m+2};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+3}{2k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{4{(-2)}^m+2\right\};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+2}{2k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{2{(-2)}^m+1\right\}+\frac{1}{2}\sum _{k=0}^m\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)};\hfill \\ \hfill \sum _{j=0}^m\sum _{k=0}^j{(-1)}^j\frac{\left(\genfrac{}{}{0pt}{}{m+1}{2k+1}\right)}{\left(\genfrac{}{}{0pt}{}{m}{j}\right)}& =\frac{m+1}{m+2}\left\{{(-2)}^m-\frac{m+1}{2}\right\}+\frac{m+3}{4}\sum _{k=0}^m\frac{1}{\left(\genfrac{}{}{0pt}{}{m}{k}\right)}.\hfill \end{array}$$

6. Concluding Comments

The telescoping approach is quite useful for establishing binomial sums. Apart from the formulae derived in the previous sections, more formulae can be found. For instance, by using the relation

$$\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}{\left(\genfrac{}{}{0pt}{}{m+k}{k}\right)}=\frac{m}{\tau m+m+\delta }({\mathrm{I}}_{k+1}+{\mathrm{I}}_k),\mathrm{with}{\mathrm{I}}_k=\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}{\left(\genfrac{}{}{0pt}{}{m+k-1}{k}\right)},$$

we can establish, through telescoping, the following formula:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}{\left(\genfrac{}{}{0pt}{}{m+k}{k}\right)}=\frac{m}{\tau m+m+\delta }+{(-1)}^m\frac{2m+1}{\tau m+m+\delta }\frac{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right)}{\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right)}.\end{array}$$

(22)

Analogously, from the relation

$$\frac{\left(\genfrac{}{}{0pt}{}{m+k}{k}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}=\frac{\tau m+\delta +1}{\tau m+m+\delta +2}({\mathcal{J}}_{k+1}+{\mathcal{J}}_k)-\frac{m+1}{\tau m+m+\delta +2}({\mathrm{J}}_{k+1}+{\mathrm{J}}_k),$$

where

$${\mathcal{J}}_k=\frac{\left(\genfrac{}{}{0pt}{}{m+k}{k}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}\mathrm{and}{\mathrm{J}}_k=\frac{\left(\genfrac{}{}{0pt}{}{m+k}{k-1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}.$$

we then obtain, by means of telescoping, the formula below:

$$\begin{array}{c}\hfill \sum _{k=0}^m{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{m+k}{k}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{k}\right)}=\frac{\tau m+\delta +1}{\tau m+m+\delta +2}+{(-1)}^m\frac{\tau m-m+\delta }{\tau m+m+\delta +2}\frac{\left(\genfrac{}{}{0pt}{}{2m+1}{m+1}\right)}{\left(\genfrac{}{}{0pt}{}{\tau m+\delta }{m+1}\right)}.\end{array}$$

(23)

In this paper, we obtained series of summation formulae concerning binomial quotients, which are difficult to calculate or cannot be calculated using classical Saalschütz, Dixon, Watson and Whipple theorems, as well as other ${}_{3}F_2$ hypergeometric series [23,25,26,27].