Analytical Relations and Statistical Estimations for Sums of Powered Integers

Abstract

Finding analytical closed-form solutions for the sums of powers of the first n positive integers is a classical problem of number theory. Analytical methods of constructing such sums produce complicated formulae of polynomials of a higher order, but they can be presented via the first two power sums. The current paper describes new presentations of the power sums and their extensions from polynomial to algebraic functions. Particularly, it shows that power sums of any higher order can be expressed just by a value of the arithmetic progression of the first power sum, or by the second power sum, or approximately by any another power sum. Regression modeling for the estimation of the powered sums is also considered, which is helpful for finding approximate values of long sums for big powers. Several problems based on the relations between sums of different powers in explicit forms are suggested for educational purposes.

1. Introduction

The sum of the first n positive integers in power p can be written as follows [1,2]:

$$S_{p,n}={\sum }_{k=1}^n{k}^p=1^p+2^p+\cdots +n^p,$$

(1)

where p is a natural number. For p = 1, it is an arithmetic progression with the formula

$$S_{1,n}={\sum }_{k=1}^n{k}^1=1+2+\cdots +n={\displaystyle \frac{n\left(n+1\right)}{2}},$$

(2)

related to the triangular numbers and known by the Pythagorean school several centuries BC. For p = 2, the closed-form solution is

$$S_{2,n}={\sum }_{k=1}^n{k}^2=1^2+2^2+\cdots +n^2={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}},$$

(3)

that was described already in around 250 BC by Archimedes. For p = 3, the sum equals

$$S_{3,n}={\sum }_{k=1}^n{k}^3=1^3+2^3+\cdots +n^3={\displaystyle \frac{n^2({n+1)}^2}{4}}.$$

(4)

As it was proved by Nicomachus (c. 60–c. 120 AD), the sum S_3,n (4) can be reduced to the square of the sum S_1,n (2):

$$S_{3,n}={\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2=S_{1,n}^2.$$

(5)

This relation is known as Nicomachus theorem [3]. An inspiring visual interpretation of the relations (2)–(5) is presented in [4].

From the 17th century and later, general solutions for sums with any power p were obtained and applied by T. Harriot, J. Faulhaber, J. Bernoulli, P. Fermat, J. Roberval, B. Pascal, C. Jacobi, L. Euler, and many others in form of the p + 1 order polynomial function by n, with the coefficients expressed via the binomial and Bernoulli numbers [1]. The basic formulae were published by German mathematician and astronomer Johann Faulhaber in 1631, also proved by Jacob Bernoulli, and later by Carl Jacobi. Additional to the first sums (2)–(4), the next several examples of sums can be presented in the factorized forms of Faulhaber polynomials [1]:

$$S_{4,n}={\sum }_{k=1}^n{k}^4=1^4+2^4+\cdots +n^4={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{30}}\left(3n^2+3n-1\right),$$

(6)

$$S_{5,n}={\sum }_{k=1}^n{k}^5=1^5+2^5+\cdots +n^5={\displaystyle \frac{n^2({n+1)}^2}{12}}(2n^2+2n-1),$$

(7)

$$S_{6,n}={\sum }_{k=1}^n{k}^6=1^6+2^6+\cdots +n^6={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{42}}\left(3n^4+6n^3-3n+1\right),$$

(8)

$$S_{7,n}={\sum }_{k=1}^n{k}^7=1^7+2^7+\cdots +n^7={\displaystyle \frac{n^2({n+1)}^2}{24}}\left(3n^4+6n^3-n^2-4n+2\right).$$

(9)

Various analytical methods of building such sums for any p-power, including matrix techniques, calculus, and the theory of numbers, have been developed in modern times, and an extensive body of the literature is devoted to this and related problems [5,6,7,8,9,10,11,12,13,14,15,16,17,18]. For example, in the work [19], it is shown how the power sums are linked to Bernoulli polynomials and which recurrent formula can be applied for finding sums of higher powers. In the work [20], it is proven that a sum of higher-order powered integers can be derived in the subsequent application of the recurrent integral relations between expressions of a lower order. Some large reviews on the history and formulae on this topic can be found in [21,22]. In the last decade, the work on these numbers and their connections to other problems has continued as well [23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38].

The current paper considers several new presentations of the power sums, relations between them, and their extensions in explicit forms. Some interesting problems based on these expressions are suggested, and they can be employed for educational purposes. Statistical modeling for the estimation of the powered sums, which is especially useful for finding approximate values of long sums with big n values, is also considered. This paper does not present a general theory but mostly concerns the relationships between the obtained forms given as special cases. These specific cases are analyzed, and the obtained results are completed by numerical methods and statistical calculations.

This paper is structured as follows. After the Introduction, Section 2 describes polynomials for the sums of positive integers in some transformations, Section 3 presents several examples of the relations between different sums, Section 4 considers regression models of the sums by their characteristics of power and length, and Section 5 summarizes the results.

2. Faulhaber Polynomials in Different Transformations

Faulhaber showed that all higher power sums can be expressed via the first two sums S_1,n (2) and S_2,n (3). Indeed, for the sums (6)–(9), they can be shown straightforwardly.

For p = 4, the expression (6) can be presented as follows:

$$S_{4,n}={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{30}}\left[3n\left(n+1\right)-1\right]={\displaystyle \frac{1}{5}}{S}_{2,n}\left(6S_{1,n}-1\right).$$

(10)

For p = 5, Formula (7) can be expressed as follows:

$$S_{5,n}={\displaystyle \frac{n^2({n+1)}^2}{12}}\left[2n\left(n+1\right)-1\right]={\displaystyle \frac{1}{3}}{S}_{1,n}^2\left(4S_{1,n}-1\right).$$

(11)

For p = 6, Formula (8) yields the following result:

$$S_{6,n}={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{42}}\left[3{n^2\left(n+1\right)}^2-3n\left(n+1\right)+1\right]={\displaystyle \frac{1}{7}}{S}_{2,n}(12S_{1,n}^2-{6S}_{1,n}+1).$$

(12)

And for p = 7, the Formula (9) can be presented as follows:

$$S_{7,n}={\displaystyle \frac{n^2({n+1)}^2}{24}}\left[3{n^2\left(n+1\right)}^2-4n\left(n+1\right)+2\right]={\displaystyle \frac{1}{3}}{S}_{1,n}^2\left(6S_{1,n}^2-{4S}_{1,n}+1\right).$$

(13)

Similar relations can be continued for the sums of higher order p. It is easy to notice that if p equals an even number, the sums have the first term $S_{2,n}$, and if p equals an odd number, the sums have the squared first term $S_{1,n}^2$. The second terms for p = 4 and p = 5, shown in parentheses in (10)–(11), are presented by the first-degree polynomials by the first sum $S_{1,n}$. For p = 6 and p = 7 in sums (12)–(13), the terms in parentheses are presented by the second-degree polynomials by $S_{1,n}$. This sequence continues for the higher power sums as well; for example, the next sums with p = 8 and p = 9 are presented via the third-degree polynomials by $S_{1,n}$, etc.

Let us consider some other presentations for the polynomials of the power sums in closed-form solutions. As it is shown in the work [39], the polynomials can be built using decompositions by the first two sums $S_{1,n}$ and $S_{2,n}$ in different powers. For example, besides (11), the formula for $S_{5,n}$ can be given as a combination of the first sum squared and the second sum squared:

$$S_{5,n}={\displaystyle \frac{3}{2}}{S}_{2,n}^2-{\displaystyle \frac{1}{2}}{S}_{1,n}^2,$$

(14)

which can also be resolved as

$$S_{2,n}^2={\displaystyle \frac{1}{3}}{S}_{3,n}+{\displaystyle \frac{2}{3}}{S}_{5,n}.$$

(15)

It is useful to note that in the last section of the article [2], this relation is given with an error (there, it is written with $S_{4,n}$ in place of $S_{3,n}$), but the correct expression is (15).

As it was shown by Faulhaber, and we can see it by the formulae given above, for the odd p values, the power sums are expressed via the first sum only, such as in (5), (11), or (13), and for the even p, the formulae also contain the multiplier $S_{2,n}$. Let us consider a new algebraic form for the powered sums.

Theorem 1. 

All power sums can be presented via the first sum $S_{1,n}$ only, and these sums are polynomials or algebraic functions.

Proof of Theorem 1. 

To prove this, consider the following transformation. The relation (2) can be represented in the quadratic equation by n as follows:

$$n^2+n-2S_{1,n}=0.$$

(16)

Solving Equation (16) with respect to a positive n yields the following root:

$$n={\displaystyle \frac{-1+\sqrt{1+8S_{1,n}}}{2}},$$

(17)

succeeding with the relation:

$$2n+1=\sqrt{8S_{1,n}+1}.$$

(18)

Using (18), we express the second power sum (3) via the first power sum:

$$S_{2,n}={\displaystyle \frac{n(n+1)(2n+1)}{6}}={\displaystyle \frac{\sqrt{8S_{1,n}+1}}{3}}{S}_{1,n}.$$

(19)

The expression (19) for $S_{2,n}$ can be substituted into all formulae for the even values of power p. Thus, not only for the odd p values are the power sums expressed via the first sum, but now, the formulae for the even p can also be represented only via $S_{1,n}$. Indeed, using the relation (19) for p = 4, substituting $S_{2,n}$ into (10) produces

$$S_{4,n}={\displaystyle \frac{1}{5}}{S}_{2,n}\left(6S_{1,n}-1\right)={\displaystyle \frac{\sqrt{8S_1,n+1}}{15}}{S}_{1,n}\left(6S_{1,n}-1\right).$$

(20)

Similarly, for p = 6, substituting $S_{2,n}$ (19) into (12) yields

$$S_{6,n}={\displaystyle \frac{1}{7}}{S}_{2,n}\left(12S_{1,n}^2-{6S}_{1,n}+1\right)={\displaystyle \frac{\sqrt{8S_{1,n}+1}}{21}}{S}_{1,n}\left(12S_{1,n}^2-{6S}_{1,n}+1\right).$$

(21)

By the same pattern, the sums for the next even powers can be obtained just as algebraic functions of only one value of the first sum $S_{1,n}$. Therefore, if a value of the arithmetic progression (2) is known, all the other higher power sums can be expressed via it as well. □

Corollary 1. 

Another representation of the power sums in more symmetric forms can be achieved in the following approach. Let us denote the square root in (18) as a new variable

$$m=\sqrt{8S_{1,n}+1}.$$

(22)

Then the expression (18) reduces to the following relations:

$$2n+1=m,n={\displaystyle \frac{m-1}{2}}, n+1={\displaystyle \frac{m+1}{2}}.$$

(23)

The first sum (2) can be rewritten via the number m as

$$S_{1,n}={\displaystyle \frac{n(n+1)}{2}}={\displaystyle \frac{m^2-1}{2^3}}.$$

(24)

With Relations (22)–(23), the second power sum (19) becomes

$$S_{2,n}={\displaystyle \frac{n(n+1)(2n+1)}{6}}={\displaystyle \frac{m(m^2-1)}{2^3\cdot 3}}.$$

(25)

The next several sums in these notations are as follows. From (5), we obtain the third sum:

$$S_{3,n}={\displaystyle \frac{n^2({n+1)}^2}{4}}={\displaystyle \frac{({m^2-1)}^2}{2^6}}.$$

(26)

From (20), we obtain the fourth sum:

$$S_{4,n}={\displaystyle \frac{1}{5}}{S}_{2,n}\left(6S_{1,n}-1\right)={\displaystyle \frac{m(m^2-1)}{2^3\cdot 3\cdot 5}}\left(3{\displaystyle \frac{m^2-1}{4}}-1\right)={\displaystyle \frac{m(m^2-1)(3m^2-7)}{2^5\cdot 3\cdot 5}}.$$

(27)

The fifth sum (11) is

$$S_{5,n}={\displaystyle \frac{1}{3}}{S}_{1,n}^2\left(4S_{1,n}-1\right)={\displaystyle \frac{({m^2-1)}^2}{2^6\cdot 3}}\left({\displaystyle \frac{m^2-1}{2}}-1\right)={\displaystyle \frac{\left({m^2-1)}^2\right(m^2-3)}{2^7\cdot 3}}.$$

(28)

The sum (21) for p = 6 becomes

$$S_{6,n}={\displaystyle \frac{1}{7}}{S}_{2,n}\left(12S_{1,n}^2-{6S}_{1,n}+1\right)={\displaystyle \frac{m(m^2-1)}{2^3\cdot 3\cdot 7}}\left(3{\displaystyle \frac{({m^2-1)}^2}{16}}-3{\displaystyle \frac{m^2-1}{4}}+1\right).$$

(29)

The sum for p = 7 equals

$$S_{7,n}={\displaystyle \frac{1}{3}}{S}_{1,n}^2\left(6S_{1,n}^2-{4S}_{1,n}+1\right)={\displaystyle \frac{({m^2-1)}^2}{2^6\cdot 3}}\left(3{\displaystyle \frac{({m^2-1)}^2}{32}}-{\displaystyle \frac{m^2-1}{2}}+1\right).$$

(30)

In a similar process, the next sums can be built. Formulae (24)–(30) present more compact forms as alternatives to the original (2)–(9) expressions. While Relations (19)–(21) correspond to the algebraic functions depending on the first sum $S_{1,n}$ and the square root of its linear form (18)–(19), the expressions (24)–(30) are given as the polynomials by the square root of the first sum (22), which by the first equation, (23), coincides with the bigger than n odd number m.

Theorem 2. 

All power sums can be presented via the second sum $S_{2,n}$ only, and these sums are algebraic functions.

Proof of Theorem 2. 

Let us represent Formula (3) of the second sum as the following equation by n:

$${2n}^3+3n^2+n-6S_{2,n}=0,$$

(31)

which corresponds to a general form of the cubic equation ${an}^3+b{n}^2+cn+d=0.$ The substitution

$$n=t-{\displaystyle \frac{b}{3a}}=t-{\displaystyle \frac{1}{2}}$$

(32)

transforms the general form into the so-called depressed cubic equation $t^3+pt+q=0$ by the new variable t, with the parameters defined as follows [40]:

$$p={\displaystyle \frac{3ac-b^2}{3a^2}}={\displaystyle \frac{6-9}{12}}={\displaystyle \frac{-1}{4}}, q={\displaystyle \frac{2b^3-9abc+27a^{2}d}{27a^3}}={\displaystyle \frac{54-54-27\ast 4\ast 6S_{2,n}}{27\ast 8}}=-3S_{2,n}.$$

(33)

The positive expression

$$\Delta ={\displaystyle \frac{q^2}{4}}+{\displaystyle \frac{p^3}{27}}={\displaystyle \frac{9S_{2,n}^2}{4}}-{\displaystyle \frac{1}{{12}^3}}>0$$

(34)

corresponds to the condition of existence of only one real root of the depressed cubic equation, which equals:

$$t=\sqrt[3]{-{\displaystyle \frac{q}{2}}+\sqrt{\Delta }}+\sqrt[3]{-{\displaystyle \frac{q}{2}}-\sqrt{\Delta }}$$

(35)

Using the expressions (32)–(34) in Formula (35) yields the unique solution of the original Equation (31):

$$n=\sqrt[3]{{\displaystyle \frac{3}{2}}{S}_{2,n}+\sqrt{{\displaystyle \frac{9}{4}}{S}_{2,n}^2-{\displaystyle \frac{1}{{12}^3}}}}+\sqrt[3]{{\displaystyle \frac{3}{2}}{S}_{2,n}-\sqrt{{\displaystyle \frac{9}{4}}{S}_{2,n}^2-{\displaystyle \frac{1}{{12}^3}}}}-{\displaystyle \frac{1}{2}}.$$

(36)

Substituting n (36) into Formulae (2), (5)–(9), or into any other sum of the powered integers expresses them via the second sum $S_{2,n}$ as algebraic functions. □

Corollary 2. 

Formula (36) can be simplified to the approximate solution: it is evident that ${\displaystyle \frac{9}{4}}{S}_{2,n}^2\gg {\displaystyle \frac{1}{{12}^3}}$ for any n value, so the small constant ${\displaystyle \frac{1}{{12}^3}}=0.000579$ can be omitted, which reduces (36) to the following simple formula:

$$n=\sqrt[3]{3S_{2,n}}-{\displaystyle \frac{1}{2}}.$$

(37)

This can be used for the approximate presentation of powered sums, especially for big n values, via the sum of the power p = 2 with good precision.

It is interesting to note that a similar expression to (37) for the first sum with p = 1 can be obtained from the simplified relation (17):

$$n={\displaystyle \frac{-1+\sqrt{1+8S_{1,n}}}{2}}={\displaystyle \frac{2\sqrt{2S_{1,n}+0.25}}{2}}-{\displaystyle \frac{1}{2}}=\sqrt{2S_{1,n}}-{\displaystyle \frac{1}{2}},$$

(38)

where the constant 0.25 is small in comparison with always a bigger value of $2S_{1,n}$, so it can be omitted. Also, for p = 3, the expression (38) with help of (5) can be represented via the sum $S_{3,n}$:

$$n=\sqrt{2S_{1,n}}-{\displaystyle \frac{1}{2}}=\sqrt{2\sqrt{S_{3,n}}}-{\displaystyle \frac{1}{2}}=\sqrt[4]{4S_{3,n}}-{\displaystyle \frac{1}{2}}.$$

(39)

Comparing Relations (37)–(39), it is easy to notice that they all correspond to the same formula

$$n=\sqrt[p+1]{(p+1)S_{p,n}}-{\displaystyle \frac{1}{2}},$$

(40)

taken for the first three values p = 1, 2, 3. Naturally, this raises a question—could this relation be true for any bigger power p?

The answer is—yes. To prove it, let us present several first sums (2)–(6) in their explicit form as regular polynomials:

$$S_{1,n}={\displaystyle \frac{n^2}{2}}+{\displaystyle \frac{n}{2}}, S_{2,n}={\displaystyle \frac{n^3}{3}}+{\displaystyle \frac{n^2}{2}}+{\displaystyle \frac{n}{6}}, S_{3,n}={\displaystyle \frac{n^4}{4}}+{\displaystyle \frac{n^3}{2}}+{\displaystyle \frac{n^2}{4}}, S_{4,n}={\displaystyle \frac{n^5}{5}}+{\displaystyle \frac{n^4}{2}}+{\displaystyle \frac{n^3}{3}}-{\displaystyle \frac{n}{30}}.$$

(41)

It is easy to notice that any sum $S_{p,n}$ in (41) is a polynomial by n of the degree p + 1, so the leading item is $n^{p+1}$ and the first coefficient is 1/(p + 1). The second item in these polynomial sums is $n^p$ with the constant coefficient ½, and the intercept equals zero. The same features are true for the sums in (7)–(9) and for higher order sums as well [1,2]. These properties are discussed in [41] (with some changes in signs defined by different limits in summation). The next items contain different coefficients defined by the Bernoulli polynomials, so the sum of powered integers (41) and other sums of any order can be presented in the following general form:

$$S_{p,n}=a_{p+1}{n}^{p+1}+a_p{n}^p+a_{p-1}{n}^{p-1}+\cdots +a_{1}n={\displaystyle \frac{n^{p+1}}{p+1}}+{\displaystyle \frac{n^p}{2}}+a_{p-1}{n}^{p-1}+\cdots +a_{1}n.$$

(42)

The main contribution to the total sum is given by the first two items, and the input from all the next items in (42) becomes negligible with big n values.

Theorem 3. 

The sum of powered integers (42) can be approximated by the first item of the depressed polynomial, and this estimate leads to the expression (40).

Proof of Theorem 3. 

Similar to the change in variable in cubic Equation (32), in the general case of polynomial (42), the substitution of the variable

$$n=t-{\displaystyle \frac{a_p}{(p+1)a_{p+1}}}=t-{\displaystyle \frac{1}{2}}$$

(43)

leads to the depressed polynomial by t without the second item and with the new coefficients b, which can be presented as follows:

$$S_{p,t}={\displaystyle \frac{t^{p+1}}{p+1}}+b_{p-1}{t}^{p-1}+\cdots +b_{1}t+b_0\backsimeq {\displaystyle \frac{t^{p+1}}{p+1}}\left(1+{\displaystyle \frac{b_{p-1}(p+1)}{t^2}}+\cdots +{\displaystyle \frac{b_1(p+1)}{t^p}}+{\displaystyle \frac{b_1(p+1)}{t^{p+1}}}\right).$$

(44)

It is useful to note that the quotient ${\displaystyle \frac{-a_p}{(p+1)a_{p+1}}}$ in (43) is related to a specific feature, where the mean value of a polynomial root equals the mean value of the locations of its critical points such as the extrema and inflection points [42]. For big n values and the correspondingly big t values (43), the sum (44) can be approximated by its first item because all the others decrease by the inverted t powers, which yields

$$S_{p,t}={\displaystyle \frac{t^{p+1}}{p+1}}$$

(45)

Returning from t to the original variable (43) transforms (45) into the following expression:

$$S_{p,n}={\displaystyle \frac{{\left(n+\frac{1}{2}\right)}^{p+1}}{p+1}}.$$

(46)

Solving Equation (46) with respect to n yields the expression (40). □

Corollary 3. 

The obtained approximation (46) can be applied for the express analysis of the powered sums, permitting their immediate evaluation. The precision of this estimate is growing with bigger n values, so for the longer sums where the simple calculation by (46) is especially useful. Concerning the expression (40), choosing the power p value and substituting n (40) into the powered sums presents any of them only via the sum $S_{p,n}$ of the taken p.

3. Several Explicit Examples of Relations Between the Sums

The considered formulae produce various interesting relations between the sums of different powers, which could be useful for instructors in educational purposes. Let us describe several such problems for exercises formulated with the sums of the p-th powers of the first n positive integers.

The expressions (10)–(11), taking (5) into account, can be rewritten as

$$S_{1,n}={\displaystyle \frac{{5S}_{4,n}}{6S_{2,n}}}+{\displaystyle \frac{1}{6}} , S_{1,n}={\displaystyle \frac{{3S}_{5,n}}{4S_{3,n}}}+{\displaystyle \frac{1}{4}}.$$

(47)

With Equation (47), a problem can be formulated in explicit form, for example, for any positive integer n, to prove the equality, or to simplify the following expression:

$$6\left(1+2+\cdots +n\right)-5{\displaystyle \frac{1^4+2^4+\cdots +n^4}{1^2+2^2+\cdots +n^2}}=1.$$

(48)

Combining both Equation (47) forms into their product yields the following expression:

$${\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1}{6}}\left(5{\displaystyle \frac{S_{4,n}}{6S_{2,n}}}+1\right)\left(3{\displaystyle \frac{S_{5,n}}{4S_{3,n}}}+1\right)}=S_{1,n},$$

(49)

which corresponds to the following problem: for any positive integer n, prove the equality

$${\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1}{6}}\left(5{\displaystyle \frac{1^4+2^4+\cdots +n^4}{1^2+2^2+\cdots +n^2}}+1\right)\left(3{\displaystyle \frac{1^5+2^5+\cdots +n^5}{1^3+2^3+\cdots +n^3}}+1\right)}=1+2+\cdots +n.$$

(50)

Excluding sum $S_{1,n}$ from Equation (47) produces another relation:

$$10{\displaystyle \frac{S_{4,n}}{S_{2,n}}}-9{\displaystyle \frac{S_{5,n}}{S_{3,n}}}=1,$$

(51)

which leads to the following problem: prove the equality, or simplify the expression at the left-hand side:

$$10{\displaystyle \frac{1^4+2^4+\cdots +n^4}{1^2+2^2+\cdots +n^2}}-9{\displaystyle \frac{1^5+2^5+\cdots +n^5}{1^3+2^3+\cdots +n^3}}=1.$$

(52)

Substituting $S_{1,n}$ (49) into the relation (18) produces another problem to prove the equality, or to simplify the given left-hand side,

$${\displaystyle \frac{1}{2}}\left(\sqrt{1+4\cdot \sqrt{{\displaystyle \frac{1}{6}}\left(5{\displaystyle \frac{1^4+2^4+\cdots +n^4}{1^2+2^2+\cdots +n^2}}+1\right)\left(3{\displaystyle \frac{1^5+2^5+\cdots +n^5}{1^3+2^3+\cdots +n^3}}+1\right)}}-1\right)=n.$$

(53)

From Formulas (11) and (13), we have the following equations:

$$3{\displaystyle \frac{S_{5,n}}{S_{3,n}}}={4S}_{1,n}-1, 3{\displaystyle \frac{S_{7,n}}{S_{3,n}}}=6S_{1,n}^2-{4S}_{1,n}+1,$$

(54)

and adding them produces the relation

$${\displaystyle \frac{S_{5,n}+S_{7,n}}{S_{3,n}}}=2S_{1,n}^2.$$

(55)

It can be suggested for proving the equality in different explicit forms, for example,

$${\displaystyle \frac{{(1}^5+2^5+\cdots +n^5)+{(1}^7+2^7+\cdots +n^7)}{{(1^3+2^3+\cdots +n^3)}^2}}=2.$$

(56)

Also, Formulas (10) and (12) can be presented as the equations

$$5{\displaystyle \frac{S_{4,n}}{S_{2,n}}}={6S}_{1,n}-1, 7{\displaystyle \frac{S_{6,n}}{S_{2,n}}}=12S_{1,n}^2-{6S}_{1,n}+1,$$

(57)

and adding them yields the relation

$${\displaystyle \frac{{5S}_{4,n}+7S_{6,n}}{S_{2,n}}}=12S_{1,n}^2,$$

(58)

which in turn can be presented as a problem of proving the following identity:

$${\displaystyle \frac{{5(1}^4+2^4+\cdots +n^4)+{7(1}^6+2^6+\cdots +n^6)}{6(1^2+2^2+\cdots +n^2)(1^3+2^3+\cdots +n^3)}}=2.$$

(59)

Equalizing Relations (55) and (58) as

$$2S_{1,n}^2={\displaystyle \frac{S_{5,n}+S_{7,n}}{S_{3,n}}}={\displaystyle \frac{{5S}_{4,n}+7S_{6,n}}{6S_{2,n}}},$$

(60)

leads to another problem to prove the equality for any given positive integer n:

$${\displaystyle \frac{6\left(1^5+2^5+\cdots +n^5\right)+6\left(1^7+2^7+\cdots +n^7\right)}{5\left(1^4+2^4+\cdots +n^4\right)+7\left(1^6+2^6+\cdots +n^6\right)}}={\displaystyle \frac{1^3+2^3+\cdots +n^3}{1^2+2^2+\cdots +n^2}}.$$

(61)

From the expression (15), we can write the next problem in the following explicit form:

$$3{(1^2+2^2+\cdots +n^2)}^2-{\left(1^1+2^1+\cdots +n^1\right)}^2=2\left(1^5+2^5+\cdots +n^5\right).$$

(62)

Equations (11) and (15) present two forms for the sum $S_{5,n}$:

$$S_{5,n}={\displaystyle \frac{4}{3}}{S}_{1,n}{S}_{3,n}-{\displaystyle \frac{1}{3}}{S}_{3,n}={\displaystyle \frac{3}{2}}{S}_{2,n}^2-{\displaystyle \frac{1}{2}}{S}_{3,n},$$

(63)

then, excluding $S_{5,n}$ leads to the relation

$$9S_{2,n}^2=8S_{1,n}^3+S_{1,n}^2,$$

(64)

which can also be obtained by squaring the relation (19). The expression (64) divided by $S_{1,n}^2$ can be simplified and presented explicitly for proving the following equality:

$${9\left({\displaystyle \frac{1^2+2^2+\cdots +n^2}{1+2+\cdots +n}}\right)}^2-8(1+2+\cdots +n)=1.$$

(65)

Combining various equations, it is easy to obtain other interesting problems for proving identities among the sums of power integers, which can be useful in classrooms.

4. Numerical Evaluation and Statistical Estimation for the Powered Sums

Additional to the analytical relations for the powered sums described above, let us consider regression modeling for their estimation. Although the straightforward calculation of sum (1) is not a hard task for a computer for given numbers of integers and power values, it is useful to obtain statistical characteristics and regression models for the dependence of different sums on the n and p values. It can be helpful for the practical express analysis of an approximate estimation of long sums with big n and big power p values. When power p is above several units, the items with bigger n values grow exponentially, and their contribution defines the main input to the total value of the sums. Statistical modeling permits us to estimate such a main portion of the sums by the biggest items.

Let us consider an approximation of the sum (1) by the corresponding integral under the assumption of continuous k values:

$$S_{p,n}={\sum }_{k=1}^n{k}^p\backsimeq {\int }_1^n{k}^{p}dk={\displaystyle \frac{k^{p+1}}{p+1}}\begin{array}{c}n\\ |\\ 1\end{array}={\displaystyle \frac{n^{p+1}-1}{p+1}}\backsimeq {\displaystyle \frac{n^{p+1}}{p+1}}.$$

(66)

The last subtraction of one in the numerator (66) is negligible in comparison with the n and p values of several units. The dependence (66) of the exponential kind can be linearized by a logarithmic transformation:

$${\ln (S}_{p,n})=\left(p+1\right)\ln \left(n\right)-\mathrm{l}\mathrm{n}\left(p+1\right),$$

(67)

which can be related to the Lambert function [43].

For statistical experiments, a data set was generated for the sums $S_{p,n}$ with the n values running from 2 to 20 and the values of p from 1 to 12. This data set is presented in

Table 1. Sums $S_{p,n}$ for p = 1, 2, …, 12, and n = 2, 3, …, 20.

p	1	2	3	4	5	6	7	8	9	10	11	12
n
2	3	5	9	17	33	65	129	257	513	1025	2049	4097
3	6	14	36	98	276	794	2316	6818	20,196	60,074	179,196	535,538
4	10	30	100	354	1300	4890	18,700	72,354	282,340	1,108,650	4,373,500	17,312,754
5	15	55	225	979	4425	20,515	96,825	462,979	2,235,465	10,874,275	53,201,625	261,453,379
6	21	91	441	2275	12,201	67,171	376,761	2,142,595	12,313,161	71,340,451	415,998,681	2,438,235,715
7	28	140	784	4676	29,008	184,820	1,200,304	7,907,396	52,666,768	353,815,700	2.393 × 109	1.628 × 1010
8	36	204	1296	8772	61,776	446,964	3,297,456	24,684,612	186,884,496	1,427,557,524	1.098 × 1010	8.4999 × 1010
9	45	285	2025	15,333	120,825	978,405	8,080,425	67,731,333	574,304,985	4,914,341,925	4.236 × 1010	3.6743 × 1011
10	55	385	3025	25,333	220,825	1,978,405	18,080,425	167,731,333	1,574,304,985	14,914,341,925	1.424 × 1011	1.3674 × 1012
11	66	506	4356	39,974	381,876	3,749,966	37,567,596	382,090,214	3,932,252,676	40,851,766,526	4.277 × 1011	4.5059 × 1012
12	78	650	6084	60,710	630,708	6,735,950	73,399,404	812,071,910	9,092,033,028	1.02769 × 1011	1.171 × 1012	1.3422 × 1013
13	91	819	8281	89,271	1,002,001	11,562,759	1.36 × 108	1.628 × 109	1.9697 × 1010	2.40628 × 1011	2.963 × 1012	3.672 × 1013
14	105	1015	11,025	127,687	1,539,825	19,092,295	2.42 × 108	3.104 × 109	4.0358 × 1010	5.29882 × 1011	7.012 × 1012	9.3414 × 1013
15	120	1240	14,400	178,312	2,299,200	30,482,920	4.12 × 108	5.666 × 109	7.8801 × 1010	1.10653 × 1012	1.566 × 1013	2.2316 × 1014
16	136	1496	18,496	243,848	3,347,776	47,260,136	6.81 × 108	9.961 × 109	1.4752 × 1011	2.20604 × 1012	3.325 × 1013	5.0464 × 1014
17	153	1785	23,409	327,369	4,767,633	71,397,705	1.09 × 109	1.694 × 1010	2.6611 × 1011	4.22204 × 1012	6.753 × 1013	1.0873 × 1015
18	171	2109	29,241	432,345	6,657,201	1.05 × 108	1.7 × 109	2.796 × 1010	4.6447 × 1011	7.79251 × 1012	1.318 × 1014	2.2441 × 1015
19	190	2470	36,100	562,666	9,133,300	1.52 × 108	2.6 × 109	4.494 × 1010	7.8716 × 1011	1.39236 × 1013	2.483 × 1014	4.4574 × 1015
20	210	2870	44,100	722,666	1.2 × 107	2.16 × 108	3.88 × 109	7.054 × 1010	1.2992 × 1012	2.41636 × 1013	4.531 × 1014	8.5534 × 1015

in the approximated form for the big p and n values, which produce huge powered sums.

Using ${y=\ln (S}_{p,n})$ as the dependent variable and items at the right-hand side (67) as two independent variables $x_1=\left(p+1\right)\ln \left(n\right)$ and $x_2=\ln (p+1)$, we can build a model of linear regression $y=a_0+a_1{x}_1+a_2{x}_2$ by the data in Table 1, with the total number of observations p × n = 12 × 20 = 240. The obtained regression model is

$${\ln (S}_{p,n})=0.941\left(p+1\right)\ln \left(n\right)-0.083\ln \left(p+1\right)-0.608,$$

(68)

with the residual standard deviation std = 0.193 and the coefficient of multiple determination R² = 0.9996, so it is of a high quality of fit. This model can be simplified by skipping the second variable, so the regression becomes

$${\ln (S}_{p,n})=0.945\left(p+1\right)\ln \left(n\right)-0.506$$

(69)

with std = 0.195 and R² = 0.9996, so this model has a high quality of fit as well. However, exponentiating the relation (68) or (69) produces more noticeable deviations from the original data from the sums.

To keep the prediction results closer to the actual values of the sums in Table 1, instead of the logarithmic transformation, it is possible to obtain a convenient-for-modeling relation by taking the first two items in polynomial (42) and dividing by the first term n^p, which yields a model:

$${\displaystyle \frac{S_{p,n}}{n^p}}={\displaystyle \frac{n}{p+1}}+{\displaystyle \frac{1}{2}}+\epsilon$$

(70)

where epsilon denotes all the smaller items in (42) corresponding to division by n in growing powers. The same model (70) can be obtained by using integration in place of summing by the more accurate Euler–Maclaurin formula [1,44], where, in place of (66), we obtain

$$S_{p,n}={\sum }_{k=1}^n{k}^p\backsimeq {\int }_1^n{k}^{p}dk+{\displaystyle \frac{1^p+n^p}{2}}={\displaystyle \frac{n^{p+1}}{p+1}}+{\displaystyle \frac{n^p}{2}}-{\displaystyle \frac{1}{p+1}}+{\displaystyle \frac{1}{2}}.$$

(71)

Dividing (71) by the term n^p and skipping the last two items quickly diminishing with bigger n values yields the main items coinciding with Formula (70).

With the function (70), we construct the linear regression model

$${\displaystyle \frac{S_{p,n}}{n^p}}=0.971{\displaystyle \frac{n}{p+1}}+0.637$$

(72)

with residual std = 0.072 and R² = 0.998 and another often used characteristic of the mean absolute error MAE = 0.051, so this model demonstrates a high quality of fit. It approximately reproduces theoretical relation (70), but with the coefficients 0.971 and 0.637 in place of 1.0 and 0.5, respectively. This model can be slightly improved by adding the predictors n and 1/(p + 1) to its mix effect, which yields

$${\displaystyle \frac{S_{p,n}}{n^p}}=-0.017n-0.698{\displaystyle \frac{1}{p+1}}+1.037{\displaystyle \frac{n}{p+1}}+0.816$$

(73)

where the quality of fit is defined by std = 0.048, R² = 0.999, and MAE = 0.036, so it is better than the previous model (72).

A few other regressions were based on the analytical approximation (46), which was tried in several versions of the transformed response variable and its possible predictors with unknown coefficients and an intercept added for enhancing the quality of fit. The model with the best characteristics was obtained using a division of relation (46) by the term ${\left(n+{\displaystyle \frac{1}{2}}\right)}^p$, which corresponds to the following dependent and independent variables:

$$y={\displaystyle \frac{S_{p,n}}{{\left(n+\frac{1}{2}\right)}^p}},x={\displaystyle \frac{n+\frac{1}{2}}{p+1}}$$

(74)

The values of quotients y (74) are shown in

Table 2. Quotients $y={\displaystyle \frac{S_{p,n}}{{(n+0.5)}^p}}$.

p	1	2	3	4	5	6	7	8	9	10	11	12
n
2	1.20	0.80	0.58	0.44	0.34	0.27	0.21	0.17	0.13	0.11	0.09	0.07
3	1.71	1.14	0.84	0.65	0.53	0.43	0.36	0.30	0.26	0.22	0.19	0.16
4	2.22	1.48	1.10	0.86	0.70	0.59	0.50	0.43	0.37	0.33	0.29	0.25
5	2.73	1.82	1.35	1.07	0.88	0.74	0.64	0.55	0.49	0.43	0.38	0.34
6	3.23	2.15	1.61	1.27	1.05	0.89	0.77	0.67	0.59	0.53	0.48	0.43
7	3.73	2.49	1.86	1.48	1.22	1.04	0.90	0.79	0.70	0.63	0.57	0.51
8	4.24	2.82	2.11	1.68	1.39	1.19	1.03	0.91	0.81	0.73	0.66	0.60
9	4.74	3.16	2.36	1.88	1.56	1.33	1.16	1.02	0.91	0.82	0.74	0.68
10	5.24	3.49	2.61	2.08	1.73	1.48	1.28	1.14	1.01	0.92	0.83	0.76
11	5.74	3.83	2.86	2.29	1.90	1.62	1.41	1.25	1.12	1.01	0.92	0.84
12	6.24	4.16	3.12	2.49	2.07	1.77	1.54	1.36	1.22	1.10	1.01	0.92
13	6.74	4.49	3.37	2.69	2.23	1.91	1.67	1.48	1.32	1.20	1.09	1.00
14	7.24	4.83	3.62	2.89	2.40	2.05	1.79	1.59	1.42	1.29	1.18	1.08
15	7.74	5.16	3.87	3.09	2.57	2.20	1.92	1.70	1.53	1.38	1.26	1.16
16	8.24	5.49	4.12	3.29	2.74	2.34	2.04	1.81	1.63	1.47	1.35	1.24
17	8.74	5.83	4.37	3.49	2.90	2.49	2.17	1.93	1.73	1.57	1.43	1.32
18	9.24	6.16	4.62	3.69	3.07	2.63	2.30	2.04	1.83	1.66	1.52	1.40
19	9.74	6.50	4.87	3.89	3.24	2.77	2.42	2.15	1.93	1.75	1.60	1.47
20	10.24	6.83	5.12	4.09	3.41	2.92	2.55	2.26	2.03	1.84	1.69	1.55

.

To check the behavior of the function y (74) by the n values, we construct graphs by the data from Table 2 and present them in Figure 1, which demonstrates an almost perfect linear dependence for any given p value.

Similarly, the behavior of the function y (74) by the p values is presented in Figure 2. It shows a hyperbolic dependence by p for the given example of n values. Thus, the dependence assumed in (46) and (74) should be of a good quality of fit.

The constructed regression model with variables (74) has the following equation:

$${\displaystyle \frac{S_{p,n}}{{\left(n+\frac{1}{2}\right)}^p}}=a_1{\displaystyle \frac{n+\frac{1}{2}}{p+1}}+a_0, a_1=1.0105, a_0=-0.0566$$

(75)

where the quality of fit is of std = 0.0237, R² = 0.9998, and MAE = 0.018, so it is the best one of all considered models. If we were to round the slope to 1 and the intercept to 0, this regression coincides with the derived relation (46). For the prediction of the sums, this regression produces values that can be found by the model represented in the following explicit form:

$$S_{p,n}={\left(n+{\displaystyle \frac{1}{2}}\right)}^p\left(1.0105{\displaystyle \frac{n+\frac{1}{2}}{p+1}}-0.0566\right).$$

(76)

This model incorporates the influence of the skipped items of the polynomials (42) and suggests the optimal estimator for the sums of powered integers in cases of various p and n values. The obtained models permit us to estimate the approximate values of the sums of powered integers, and they are convenient, especially in the cases of big n and p numbers. The prediction intervals can be defined as a predicted value plus–minus a critical t-statistics value (for example, t = 1.96 for 95% of the confidence level or a level of significance of 5%) multiplied by the error estimate [45], and more detail on the interval prediction by regression models can be found in any book on regression analysis.

Table 3. Quotients of the predicted values to the exact values of the sums $S_{p,n}$.

p	1	2	3	4	5	6	7	8	9	10	11	12	mean	std
n
2	0.99	0.97	0.99	1.02	1.07	1.13	1.21	1.31	1.44	1.59	1.77	1.97	1.29	0.34
3	0.99	0.97	0.97	0.99	1.00	1.03	1.06	1.10	1.15	1.20	1.27	1.34	1.09	0.12
4	0.99	0.97	0.97	0.98	0.98	1.00	1.01	1.03	1.05	1.08	1.12	1.15	1.03	0.06
5	0.99	0.98	0.97	0.98	0.98	0.98	0.99	1.00	1.02	1.03	1.05	1.07	1.00	0.03
6	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.99	1.00	1.01	1.02	1.03	0.99	0.02
7	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.99	1.00	1.00	1.01	0.99	0.01
8	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.99	0.99	1.00	0.98	0.01
9	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.99	0.99	0.98	0.01
10	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.99	0.98	0.00
11	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.00
12	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.00
13	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.00
14	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
15	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
16	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
17	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
18	0.99	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
19	0.99	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.01
20	1.00	0.99	0.99	0.99	0.99	0.98	0.98	0.98	0.98	0.98	0.98	0.98	0.99	0.01
mean	0.99	0.98	0.98	0.98	0.99	0.99	1.00	1.01	1.02	1.03	1.05	1.07	1.01
std	0.00	0.01	0.01	0.01	0.02	0.04	0.05	0.08	0.11	0.14	0.19	0.24		0.11

presents quotients of the predicted values (76) to the original exact values of the sums from Table 1. The last two columns show the means and the standard deviation of the values in each row (so for each n value). Similarly, the last two rows demonstrate the mean and std of the values in each column for each p value. The last two values in the south-east corner of the table present the mean and std across the deviations of all data. The quotients are close to one, and the std values are rather small, so the approximation (76) is very good, especially for the big n values, for which the prediction by models is especially useful. All characteristics of the regressions’ quality indicate that the models are valuable, but of course, they give just approximations of the powered sums, so they can serve for a quick estimation, while finding an exact result requires the calculation of the sums by their definition or the analytical closed-form solutions described above.

5. Summary

This paper considers a classical problem of the number theory about the sums of integers in different powers and the relations between them. Some new transformations are described, particularly with the extension of polynomials to algebraic functions. It is shown that by using these algebraic functions, the power sums of any higher order can be expressed just by a value of the arithmetic progression of the first power sum or any other taken sum. Problems based on the relations between sums of different powers in explicit forms are suggested for educational purposes. Statistical evaluations and regression modeling for an approximation of the sums of powered integers are performed as well, and such models have good precision, especially for long sums. Future research can be focused on other possible transformations of polynomials of sums of powered integers in different extended forms with relations between them both in analytical and statistical models.