A special case of triple product identity

Question

Can any one guide me how can I prove these identities??

$$\sum_{n=-\infty}^\infty (-1)^nq^{(6n+1)^2}=q \prod_{n=1}^\infty(1-q^{24n})$$

$$\sum_{n=-\infty}^\infty (4n+1)q^{(4n+1)^2}=q \prod_{n=1}^\infty(1-q^{8n})^3$$

Answer 1

I'll show how to do the first problem. Consider the Jacobi triple product identity

$$\sum_{n = -\infty}^\infty z^nq^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1}).$$

Replacing $q$ by $q^{36}$, then setting $z = -q^{12}$, we have

\begin{align}\sum_{n = -\infty}^\infty (-1)^n q^{36n^2 + 12n} &= \prod_{n = 1}^\infty (1 - q^{72n})(1 - q^{72n-24})(1 - q^{72n - 48})\\ &= \prod_{n = 1}^\infty (1 - (q^{24})^{3n})(1 - (q^{24})^{3n-1})(1 - (q^{24})^{3n-2})\\ &= \prod_{n = 1}^\infty (1 - q^{24n}). \end{align}

Multiplying by $q$, we obtain

$$\sum_{n = -\infty}^\infty (-1)^n q^{(6n+1)^2} = q\prod_{n = 1}^\infty (1 - q^{24n}).$$