A \(q\)-analogue of Faulhaber's formula for sums of powers (Q2570988)

Almost \(400\) years ago, J. Faulhaber computed the sums \[ S_{m,n}=\sum_{k=1}^n k^m \] for all \(m\leq 17\). For example, \[ S_{3,n}=1^3+2^3+\cdots+n^3=\binom{n+1}{2}^2.\tag{cubes} \] Oddly enough, only recently has the problem of computing \(q\)-analogues of sums of powers attracted the attention of mathematicians. Two years ago \textit{K. C. Garrett} and \textit{K. Hummel} [A combinatorial proof of the sum of \(q\)-cubes, Electron. J. Comb. 11, Research paper R9 (2004; Zbl 1050.05012)] found -- by combinatorial means -- a \(q\)-analogue of the sum of cubes. Unfortunately, their result lacked the elegance of cubes, and Garrett and Hummel asked for a more appealing \(q\)-version. This was provided by the reviewer [On the \(q\)-analogue of the sum of cubes, Electron. J. Comb. 11, Research paper N13, electronic only (2004; Zbl 1071.05010)] as \[ \sum_{k=1}^n\frac{(1-q^{2k})(1-q^k)^2}{(1-q^2)(1-q)^2} \, q^{2n-2k}= \frac{(1-q^n)^2(1-q^{n+1})^2}{(1-q)^2(1-q^2)^2}. \] This result was further generalized by \textit{M. Schlosser} [\(q\)-analogues of the sums of consecutive integers, squares, cubes, quarts and quints, Electron. J. Comb. 11, Research paper R71, electronic only (2004; Zbl 1064.33014)] who computed the \(q\)-Faulhaber sums \[ S_{m,n}(q)=\sum_{k=1}^n \frac{1-q^{2k}}{1-q^2} \biggl(\frac{1-q^k}{1-q}\biggr)^{m-1}q^{(m+1)(n-k)/2}\tag{qS} \] for all \(m\leq 5\) using basic hypergeometric series techniques. In the present paper Schlosser's programme has been completed, providing the following closed form expression for \(S_{m,n}(q)\) for all \(m\): \[ \begin{aligned} S_{2m+1,n}(q)&=\sum_{k=0}^m(-q^n)^k P_{m,k}(q)\, \frac{[(1-q^n)(1-q^{n+1})]^{m-k+1}} {(1-q^2)(1-q)^{2m-3k}\prod_{i=0}^k (1-q^{m-i+1})} \\ S_{2m,n}(q)&=\sum_{k=0}^m(-q^n)^k Q_{m,k}(q^{1/2})\, \frac{(1-q^{1/2})^k(1-q^{n+1/2})[(1-q^n)(1-q^{n+1})]^{m-k}} {(1-q^2)(1-q)^{2m-2k-1}\prod_{i=0}^k (1-q^{m-i+1/2})} \end{aligned} \] where \(P_{m,k}(q)\) and \(Q_{m,k}(q)\) are polynomials given by \[ \begin{multlined} P_{m,k}(q)=\frac{\prod_{j=0}^k(1-q^{m-j+1})}{(1-q)^{3k}} \sum_{j=0}^k \frac{(-1)^{k-j}}{1-q^{m-j+1}} \biggl[\binom{2m}{j}-\binom{2m}{j-2}\biggr] \\ \times \sum_{i=0}^{k-j}\frac{m-j+1}{m-k+1}\binom{m-k+i}{i} \binom{m-i-j}{k-i-j}q^{k-i-j} \end{multlined} \] and \[ \begin{multlined} Q_{m,k}(q)=\frac{\prod_{j=0}^k(1-q^{2m-2j+1})}{(1-q)^k(1-q^2)^{2k}} \sum_{j=0}^k \frac{(-1)^{k-j}}{1-q^{2m-2j+1}} \biggl[\binom{2m-1}{j}-\binom{2m-1}{j-2}\biggr] \\ \times \sum_{i=0}^{k-j} \binom{m-k+i}{i}\biggl[ \binom{m-i-j}{k-i-j}q^{2k-2i-2j}+ \binom{m-i-j-1}{k-i-j-1}q^{2k-2i-2j-1}\biggr]. \end{multlined} \] A similar result is proved for the \(q\)-analogue of the alternating sum of \(m\)th powers \[ \sum_{k=1}^n (-1)^{n-k}k^m. \] The authors also announce a forthcoming paper in which a combinatorial interpretation of the coefficients of \(P_{m,k}(q)\) and \(Q_{m,k}(q)\) is given, implying that these coefficients are in fact nonnegative.