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I need to simplify this expression:$$\sum_{k=0}^n{n \choose k}^2(-1)^k$$ I know that $$\sum_k^n {n \choose k}^2 = {2n \choose n}$$ but that $(-1)^k$ spoils everything. I tried writing it like this:
$$\sum_{k=0}^n{n \choose k}^2(-1)^k = {2n \choose n} - 2 \left({n \choose 1 }+ {n \choose 3} + \dots + {n \choose n} \right) $$
for odd $n$ and up to ${n \choose n-1}$ for even $n$ - but I am not able to sum the expression in parentheses.

Also, I tried to use this relation: $$(1-x)^n(1+x)^n = (1-x^2)^n$$ Now, using the binomial coefficient and substituting n = 1 I got:
$$\sum_{k=0}^n {n \choose k}(-1)^k = \left(\sum_{k=0}^n {n \choose k}\right) \left( \sum_{k=0}^n{n \choose k}(-1)^k\right)$$ Which can be distributed according to Cauchy theorem:
$$= \sum_{k=0}^n \sum_{i=0}^k{k \choose i}^2(-1)^k$$
But this is still to no avail...
Could you tell me if any of my methods can lead to solving this problem? If not, what other things can I try to solve it?

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1 Answer 1

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You are nearly there; your sum is $$\sum_{k=0}^n(-1)^k\binom nk\binom{n}{n-k}$$ which is the coefficient of $x^n$ in $(1-x)^n(1+x)^n=(1-x^2)^n$. This is zero whenever $n$ is odd, and when $n$ is even it is $$(-1)^{n/2}\binom n{n/2}.$$

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