# Deducing the Gaussian Formula

How would one prove the following formula: For all $n \in \mathbb{N}$, $$\sum_{m=0}^n (-1)^m \binom{n}{m}_q = \begin{cases} \displaystyle \mathop{\prod_{k \text{ odd}}}_{1 \le k\leq n} (1-q^k), & \text{if } n \text{ is even}\\[30pt]0, &\text{if } n \text{ is odd}. \end{cases}$$

I know the odd case follows swiftly from the symmetry property of $q$-binomial coefficients, but I am having trouble working out the even case. Any help would be appreciated!

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The recurrence of the $q-$ binomial coefficients gives $\binom{2n}{m}_q$= $q^m \binom{2n-1}{m}_q+ \binom{2n-1}{m-1}_q$ and therefore $\sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =\sum_{m=0}^{2n-1} (-1)^m q^m\binom{2n-1}{m}_q .$ Formula $(1-q^m) \binom{2n-1}{m}_q=(1-q^{2n-1}) \binom{2n-2}{m-1}_q$ implies therefore $\sum_{m=0}^{2n} (-1)^m \binom{2n}{m}_q =(1-q^{2n-1})\sum_{m=0}^{2n-2} (-1)^{m-1}\binom{2n-2}{m-1}_q .$

Edit

A more elegant proof is the following: Let $r(n)=\sum_{m=0}^n (-1)^m \binom{n}{m}_q.$

Consider the $q-$ exponential function $e(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{(1 - q) \cdots (1 - {q^n})}}} .$

Comparing coefficients gives $\sum\limits_{n = 0}^\infty {\frac{{r(n)}}{{(1 - q) \cdots (1 - {q^n})}}} = e( - x)e(x).$

Now $e(x) = \frac{1}{{(1 - x)(1 - qx)(1 - {q^2}x) \cdots }}$ and therefore $e( - x)e(x) = \frac{1}{{(1 - {x^2})(1 - {q^2}{x^2})(1 - {q^4}{x^2}) \cdots }} = \sum\limits_{}^{} {\frac{{{x^{2n}}}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}}.}$

Comparing coefficients of the last two series gives

$r(2n) = \frac{{(1 - q)(1 - {q^2}) \cdots (1 - {q^{2n}})}}{{(1 - {q^2})(1 - {q^4}) \cdots (1 - {q^{2n}})}} = (1 - q)(1 - {q^3}) \cdots (1 - {q^{2n - 1}}).$

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