Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.