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I'm attempting to prove that $$\sum_{k=0}^n(-1)^k{n \choose k}^2=\begin{cases}0&\text{if $n$ is odd}\\(-1)^m{2m \choose m}&\text{if $n$ = 2m}\end{cases}$$ via induction. Here's what I have so far:
Base Case (odd): $n=1$ $$\sum_{k=0}^1(-1)^k{1 \choose k}^2=0$$ Base Case (even): $n=2$ $$\sum_{k=0}^2(-1)^k{2 \choose k}^2=-2$$ Hypothesis: $n=l$ $$\sum_{k=0}^l(-1)^k{l \choose k}^2=\begin{cases}0&\text{if $n$ is odd}\\(-1)^m{2m \choose m}&\text{if $n$ = 2m}\end{cases}$$ Inductive Step: $n=l+1$ $$\sum_{k=0}^{l+1}(-1)^k{l+1 \choose k}^2$$ I know that the goal is to get this sum into terms of the hypothesis plus the $(l+1)^{th}$ term, but I'm not sure how. I've tried manipulating various identities, but so far I'm not coming up with anything workable. Thanks in advance for any help.

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marked as duplicate by Felix Marin, Claude Leibovici, Namaste, projectilemotion, Daniel W. Farlow Mar 29 '17 at 19:39

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  • $\begingroup$ none are induction proofs? $\endgroup$ – miniparser Mar 31 '17 at 21:30
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$\sum_{k=0}^n(-1)^k{n \choose k}^2 = \sum_{k=0}^n(-1)^k{n \choose k}\binom{n}{n-k}$ is the coefficient of $x^n$ in $(1-x)^n(1+x)^n=(1-x^2)^n$ which only contains even powers of $x$ in its expansion, so the coefficient of $x^n$ there is $0$ when $n$ is odd and $(-1)^{n/2}\binom{n}{n/2}$ when $n$ is even.

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