A q-Analogue of Faulhaber's Formula for Sums of Powers

Abstract

Let $$S_{m,n}(q):=\sum _{k=1}^n\frac{1-q^{2k}}{1-q^2}(\frac{1-q^k}{1-q}{)}^{m-1}{q}^{\frac{m+1}{2}(n-k)}.$$ Generalizing the formulas of Warnaar and Schlosser, we prove that there exist polynomials $P_{m,k}(q)\in \mathbb{Z}[q]$ such that $$S_{2m+1,n}(q)=\sum _{k=0}^m(-1{)}^k{P}_{m,k}(q)\frac{(1-q^n{)}^{m+1-k}(1-q^{n+1}{)}^{m+1-k}{q}^{kn}}{(1-q^2)(1-q{)}^{2m-3k}{\prod }_{i=0}^k(1-q^{m+1-i})},$$ and solve a problem raised by Schlosser. We also show that there is a similar formula for the following $q$-analogue of alternating sums of powers: $$T_{m,n}(q):=\sum _{k=1}^n(-1{)}^{n-k}(\frac{1-q^k}{1-q}{)}^m{q}^{\frac{m}{2}(n-k)}.$$

arXiv comment19 pages

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• q-analogues of the sums of powers of consecutive integers