# Finite version special case Jacobi triple product formula

In this paper, Shanks uses the following formula: $$\sum_{s=0}^{n-1}q^{s(2n+1)} \times \left( \prod_{k=s+1}^{n} \dfrac{1-q^{2k}}{1-q^{2k-1}}\right) = \sum_{s=1}^{2n} q^{\frac{s(s-1)}{2}}$$ to get a one line computation of the Gauss sum: $$\sum_{k=0}^{m-1} e^{\frac{2ik^2 \pi}{m}}$$ in the odd case ($$m=4n+1$$ or $$m=4n+3$$).

He proves this relation by induction which is fine with me. I was wondering however if there is a combinatorial interpretation of this formula? Perhaps using the $$q$$-binomial Theorem and the inverse of the $$q$$-Pascal matrix?

Apart from its use in Gauss sum computations, this formula can be seen as a finite version of a special case of Jacobi triple product identity. Indeed, if we identify on the left and right handsides the terms having degree $$\leq 2n$$ and letting $$n$$ go to infinity, we get:

$$\prod_{k=1}^{+ \infty} \dfrac{1-q^{2k}}{1-q^{2k-1}} = \sum_{k=1}^{+ \infty} q^{\frac{k(k-1)}{2}}.$$

• This does not answer your question, but it might motivate an analogous answer: MathOverflow.net/q/155114 . I had a problem involving truncated sums which yielded to a nice analytical transform posted by Aaron Meyerowitz. Gerhard "Maybe The Answer Is Analytic?" Paseman, 2019.10.22. – Gerhard Paseman Oct 22 '19 at 16:12
• I am also taking a "simple" analytic proof if there is one. – Libli Oct 22 '19 at 16:25