This is a question of an assignment I am solving and I am unable to think about the case when n=odd.
Let n be a positive integer. Prove that $\sum_{k=0}^n (-1)^k {n\choose k}^2 = 0 $ if n is odd .
For n = even integer I was able to prove the required result ( whose statement I am not mentioning) but for this case I am unable to think about it.
Can someone please tell how to prove it.
Hint: Expand both sides of $$ (1 - x)^n (1 + x)^n = (1 - x^2 )^n $$ and note that $\binom{n}{k}=\binom{n}{n-k}$.