A New Formula for the Sum of Integer Powers Based on Elementary Matrix Theory

#A98 INTEGERS 25 (2025) A NEW FORMULA FOR THE SUM OF INTEGER POWERS BASED ON ELEMENTARY MATRIX THEORY Peiran Ding Wuxi Dipont School of Arts and Science, Wuxi, Jiangsu, China [email protected] Yanjun Zhu Wuxi Dipont School of Arts and Science, Wuxi, Jiangsu, China [email protected] Lei Wang1 Scholastic Excellence Research Center, Wuxi Dipont School of Arts and Science, Wuxi, Jiangsu, China [email protected] Received: 7/16/24, Revised: 7/16/25, Accepted: 10/7/25, Published: 11/5/25 Abstract In this study, we present a novel formula for the sum of the k-th powers of the first npositive integers, denoted as S(n, k). Unlike existing formulations that employ binomial expansions, generating functions, or make use of special numbers such as Bernoulli or Stirling numbers, our approach leverages fundamental principles from elementary matrix theory. Additionally, we demonstrate that for very large values of n, the derived formula can be partially simplified. By contrasting our findings with a previously established Bernoulli-type formula, we ultimately derive a new combinatorial identity. 1. Introduction Our investigation begins with the well-established expression for S(n, k), defined as the sum of the k-th powers of the first npositive integers: S(n, k) = 1k+ 2k+· · · +nk. The formulas for S(n, k) have been the subject of extensive study over several centuries [3, 6, 9, 11]. In the past decade, numerous researchers have continued to DOI: 10.5281/zenodo.17535263 1corresponding author INTEGERS: 25 (2025) 2 explore the derivation of new formulas as well as the proof of existing ones from various methodologies [2, 4, 5, 7, 8, 10]. A prominent formula in this domain is derived from Pascal’s identity in conjunction with the binomial theorem, given by S(n, k) = nk+1 k+ 1 + k−1 X r=0 k r(−1)k−r+1 k−r+ 1 S(n, r).(1) This expression reveals a clear recurrence relation between S(n, k) and S(n, r) (see Equation (1)), thus, the computation of S(n, k) necessitates the prior determination of S(n, r). Comprehensive analyses and proofs concerning this formula can be found in references [7, 8]. Subsequently, several researchers have sought to develop new formulas that eliminate these recurrence relations, introducing novel elements such as Bernoulli numbers [3, 9], Stirling numbers [5, 10], and hyperharmonic numbers [4]. Some of these formulations are expressed in the following equations: S(n, k) = nk+1 k+ 1 +nk 2+ k X i=2 Bi ik i−1nk−i+1,(2) S(n, k) = k X r=0 r!k rn+ 1 r+ 1,(3) S(n, k) = (−1)k+1 k+ 1 k+1 X j=1 (−1)jj!k+ 1 jH(n) j+1 −1 j+ 1.(4) In these equations, Birepresents the Bernoulli numbers, k rdenotes the Stirling numbers of the second kind, and H(n) j+1 signifies the (j+1)-th hyperharmonic number. Detailed discussions and proofs of these formulas are available in the cited references. A close examination of the derivation processes for formulas (1)-(4) reveals that they predominantly rely on binomial expansions, generating functions, or the utilization of special numbers. In contrast, our study derives a new formula for S(n, k) through the lens of matrix theory, employing several key results related to matrix traces. Furthermore, we establish a new combinatorial identity by juxtaposing our simplified formula with a previously established Bernoulli-type expression. 2. Preliminaries 2.1. Relationship Between S(n, k) and the Trace of Matrices It is well-established that real-valued n×ndiagonal matrices contain nonzero entries exclusively on their main diagonal, with all off-diagonal elements equal to INTEGERS: 25 (2025) 3 zero. Consider a diagonal matrix Awhose main diagonal consists of the elements 1,2, . . . , n: A=     1 0 · · · 0 0 2 · · · 0 . . .. . ..... . . 0 0 · · · n      . Utilizing the fundamental rules of matrix multiplication, we find that the k-th power of matrix Ais given by Ak=     1k0· · · 0 0 2k· · · 0 . . .. . ..... . . 0 0 · · · nk      . From the definition of the trace of a matrix, we can derive the following relationship: Tr Ak= 1k+ 2k+· · · +nk=S(n, k). This relation indicates that the analysis of the formula for S(n, k) can be transformed into an investigation of the trace Tr Ak. For small values of k, such as k= 1,2,3, the traces are well-known: Tr A1= 11+ 21+· · · +n1=n(n+ 1) 2, Tr A2= 12+ 22+· · · +n2=n(n+ 1)(2n+ 1) 6, Tr A3= 13+ 23+· · · +n3=n2(n+ 1)2 4. However, for larger values of k(k > 3), obtaining a general formula becomes increasingly complex. Insights from a previously established theorem discussed in Section 2.2, which presents two formulas characterizing the traces of 2 ×2 realvalued matrices raised to the k-th power, will be instrumental in advancing our study. 2.2. A Previously Established Theorem Theorem 1. For any positive integer mand for 2×2real-valued matrices M, the following relations hold: Tr M2m= (2m) m X r=0 (−1)r r! (2m−r−1)! (2m−2r)! [Det(M)]r[Tr(M)]2m−2r,(5) and Tr M2m+1= (2m+ 1) m X r=0 (−1)r r! (2m−r)! (2m+ 1 −2r)! [Det(M)]r[Tr(M)]2m+1−2r,(6) INTEGERS: 25 (2025) 4 where Tr(·)and Det(·)denote the trace and determinant of the corresponding matrices, respectively. Proof. The proof of the formulas presented in Theorem 1 can be established using the method of mathematical induction. A comprehensive explanation of the proof can be found in reference [1]. 2.3. Decomposition of S(n, k) To utilize the conclusions derived from Theorem 1 in formulating a new expression for Tr Ak, it is necessary to restrict the dimension of matrix Ato 2 ×2 or to decompose the n×nmatrix Ainto smaller 2 ×2 matrices. Consequently, since Tr Ak=S(n, k), we can express S(n, k) as the sum of the traces of these decomposed 2 ×2 matrices. Lemma 1. For any positive integer kand even positive integer n,S(n, k)can be expressed as the sum of ⌊n/2⌋traces of 2×2real-valued matrices, S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s, where s=⌊n/2⌋, and Mi=i0 0 2s+ 1 −iare 2×2real-valued matrices for 1⩽i⩽s. Proof. Let n= 2s. Thus, we can rewrite S(n, k) as S(n, k) = 1k+ 2k+· · · + (2s)k. The 2sterms in S(n, k) can be grouped into s=⌊n/2⌋pairs. For each pair, such as ik+ (2s+ 1 −i)k, we have Mi=i0 0 2s+ 1 −i, leading to the relation ik+ (2s+ 1 −i)k= Tr Mk i. Since there are different Mis, we conclude that S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s, where M1, M2, . . . , Msare 2 ×2 real-valued matrices for all 1 ⩽i⩽s. So, Lemma 1 holds. Lemma 2. For any positive integer kand odd positive integer n,S(n, k)can be expressed as the sum of ⌊n/2⌋traces of 2×2real-valued matrices, plus nk, S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s+nk, where s=⌊n/2⌋, and Miare 2×2real-valued matrices for 1⩽i⩽s. INTEGERS: 25 (2025) 5 Proof. Since nis odd, we can set n= 2s+ 1, where sis a positive integer. Thus, we can rewrite S(n, k) as S(n, k)=1k+ 2k+· · · + (2s)k+ (2s+ 1)k. By focusing on the first 2sterms, we can apply a procedure analogous to that used in the proof of Lemma 1. Therefore, we establish that S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s+nk, where M1, M2, . . . , Msare 2×2 real-valued matrices for all 1 ⩽i⩽s. Thus, Lemma 2 holds. 2.4. Definition of the Function W(n, r) In this section, we introduce the function W(n, r), which is a crucial component in our derived formula. The function is defined as W(n, r) = n X i=1  i n+ 1r1−i n+ 1r. Letting xi=i n+1 and f(xi) = xr i(1 −xi)r, we note that since 1 ⩽i⩽n, it follows that 0 < xi<1. The function f(x) = xr(1 −x)ris characterized by a symmetrical shape around x= 0.5. Consequently, when nis even, W(n, r) can be rewritten as W(n, r) = f(x1) + f(x2) + · · · +f(xn/2) + f(xn/2+1) + · · · +f(xn−1) + f(xn). From the definition of xi, we can deduce the following relationships: x1+xn=1 n+ 1 +n n+ 1 =1 + n n+ 1 = 1 = 2 ×0.5 implies f(x1) = f(xn), x2+xn−1=2 n+ 1 +n−1 n+ 1 =1 + n n+ 1 = 1 = 2 ×0.5 implies f(x2) = f(xn−1), . . . xn/2+xn/2+1 =n/2 n+ 1 +n/2+1 n+ 1 =n+ 1 n+ 1 = 1 = 2 ×0.5 implies f(xn/2) = f(xn/2+1). Thus, we conclude that n/2 X i=1  i n+ 1r1−i n+ 1r=W(n, r) 2. INTEGERS: 25 (2025) 6 3. Main results 3.1. A New Formula for S(n, k) In this section, we derive our primary formula for S(n, k). Theorem 2. For any positive integers kand n, the quantity S(n, k)can be expressed as S(n, k) = k(n+ 1)k 2 ⌊k/2⌋ X r=0 (−1)r k−rk−r rW(n, r),(7) where W(n, r)is the function defined in Section 2.4. Proof. We consider two cases. Case I: nis even. In this case, based on Lemma 1, we have S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s, where Miare the 2 ×2 matrices defined previously. If kis even, let k= 2m. Utilizing the formula from Equation (5) alongside the values of Det(Mi) and Tr(Mi), we obtain Tr M2m i= (2m) m X r=0 (−1)r r! (2m−r−1)! (2m−2r)! [Det(Mi)]r[Tr(Mi)]2m−2r = (2m) m X r=0 (−1)r r! (2m−r−1)! (2m−2r)! [i(2s+ 1 −i)]r[2s+ 1]2m−2r = (2m)(2s+ 1)2m m X r=0 (−1)r r! (2m−r−1)! (2m−2r)! [i(2s+ 1 −i)]r [2s+ 1]2r = (2m)(2s+ 1)2m m X r=0 (−1)r 2m−r2m−r r i 2s+ 1r1−i 2s+ 1r. Thus, we can express S(n, k) as S(n, k) = (2m)(2s+ 1)2m m X r=0 (−1)r 2m−r2m−r rs X i=1  i 2s+ 1r1−i 2s+ 1r. Substituting n= 2sand k= 2m, we derive S(n, k) = k(n+ 1)k ⌊k/2⌋ X r=0 (−1)r k−rk−r rn/2 X i=1  i n+ 1r1−i n+ 1r. INTEGERS: 25 (2025) 7 Thus, we conclude that S(n, k) = k(n+ 1)k 2 ⌊k/2⌋ X r=0 (−1)r k−rk−r rW(n, r). If kis odd, let k= 2m+1. Using the formula from Equation (6) with the relevant values of Det(Mi) and Tr(Mi), we obtain Tr M2m+1 i= (2m+ 1) m X r=0 (−1)r r! (2m−r)! (2m+ 1 −2r)! [Det(Mi)]r[Tr(Mi)]2m+1−2r = (2m+ 1) m X r=0 (−1)r r! (2m−r)! (2m+ 1 −2r)![i(2s+ 1 −i)]r[2s+ 1]2m+1−2r = (2m+ 1)(2s+ 1)2m+1 m X r=0 (−1)r r! (2m−r)! (2m+ 1 −2r)! [i(2s+ 1 −i)]r [2s+ 1]2r = (2m+ 1)(2s+ 1)2m+1 m X r=0 (−1)r 2m+ 1 −r2m+ 1 −r r · i 2s+ 1r1−i 2s+ 1r. Thus, we express S(n, k) as S(n, k) = (2m+ 1)(2s+ 1)2m+1 m X r=0 (−1)r 2m+ 1 −r2m+ 1 −r r · s X i=1  i 2s+ 1r1−i 2s+ 1r. Substituting n= 2s,k= 2m+ 1, we find S(n, k) = k(n+ 1)k ⌊k/2⌋ X r=0 (−1)r k−rk−r rn/2 X i=1  i n+ 1r1−i n+ 1r. Thus, we conclude that S(n, k) = k(n+ 1)k 2 ⌊k/2⌋ X r=0 (−1)r k−rk−r rW(n, r). Case II: nis odd. In this case, based on Lemma 2, we have S(n, k) = Tr Mk 1+ Tr Mk 2+· · · + Tr Mk s+nk. INTEGERS: 25 (2025) 8 Utilizing similar arguments and analytical strategies as in Case I, we can show that when nis odd, S(n, k) can be expressed as S(n, k) = knk 2 ⌊k/2⌋ X r=0 (−1)r k−rk−r rW(n−1, r) + nk.(8) The detailed proofs of this portion are provided in the Appendix. Next, we demonstrate that Equations (7) and (8) are equivalent. From the definition of S(n, k), we have S(n, k) = S(n−1, k) + nkfor all positive integers n. In particular, if nis odd, then n−1 is even, allowing us to apply Equation (7) to arrive at Equation (8). Consequently, Equation (7) is valid for odd values of n. Thus, for any positive integers kand n, we conclude that S(n, k) = k(n+ 1)k 2 ⌊k/2⌋ X r=0 (−1)r k−rk−r rW(n, r). Therefore, Theorem 2 is established. 3.2. Asymptotic Formula for S(n, k) as nApproaches Infinity In Section 2.4, we defined the function W(n, r). Here, we extend our discussion of this function by examining the asymptotic behavior as napproaches infinity. Recall the expression for W(n, r), given by W(n, r) = n X i=1  i n+ 1r1−i n+ 1r. Letting xi=i n+1 , we can express this as W(n, r) = n X i=1 f(xi), where f(xi) = xr i(1−xi)r. Notably, since x0=0 n+1 = 0 and 1−xn+1 = 1−n+1 n+1 = 0, we find that f(x0) = 0 and f(xn+1) = 0. Thus, we can rewrite W(n, r) as W(n, r) = n+1 X i=0 f(xi) = (n+ 1) · n+1 X i=0 f(xi)1 n+ 1, here, 1 n+1 serves as the width of each subinterval, ∆xi, and as nbecomes very large, ∆xi≈dx. By the definition of the definite integral, we can thus approximate W(n, r)≈(n+ 1) Z1 0 f(x) dx= (n+ 1) Z1 0 xr(1 −x)rdx. INTEGERS: 25 (2025) 9 Definition 1. (Beta function). The following function B(a, b) is defined as the Beta function, if it satisfies B(a, b) = Z1 0 xa−1(1 −x)b−1dx, where a,bare real numbers. Definition 2. (Gamma function). The following function Γ(n) is defined as the Gamma function, if it satisfies Γ(n) = Z∞ 0 xn−1e−xdx= (n−1)!, where nis a positive integer. Lemma 3. The following holds for all positive integers r: Z1 0 xr(1 −x)rdx=1 (2r+ 1)2r r. Proof. By Definition 1, we have B(a, b) = Z1 0 xa−1(1 −x)b−1dx. Thus, it follows that Z1 0 xr(1 −x)rdx=B(r+ 1, r + 1). By Definition 2, we have Γ(n)=(n−1)!. Employing the relationship between the Beta function and the Gamma function [12], we obtain B(a, b) = Γ(a)Γ(b) Γ(a+b). This leads to B(r+ 1, r + 1) = Γ(r+ 1)Γ(r+ 1) Γ(2r+ 2) . Finally, we derive B(r+ 1, r + 1) = r!·r! (2r+ 1)! =r!·r! (2r)! 1 (2r+ 1) =1 (2r+ 1)2r r. Thus, Lemma 3 is established.