TITLE: Show that $\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$
QUESTION [18 upvotes]: How can I prove the identity: 
$$
\sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}?
$$
Maybe, can we expand
$$
f(x)=(1+x)^{2n}?
$$
Thank you.

REPLY [2 votes]: This is simply
$$\sum_{k=0}^{2n} {2n\choose k}^2 (-1)^k
=\sum_{k=0}^{2n} {2n\choose k} {2n\choose 2n-k} (-1)^k
\\ = [z^{2n}] (1+z)^{2n} \sum_{k=0}^{2n} {2n\choose k} z^k (-1)^k
\\ = [z^{2n}] (1+z)^{2n} (1-z)^{2n}
= [z^{2n}] (1-z^2)^{2n} = [z^n] (1-z)^{2n} \\ = (-1)^n {2n\choose n}.$$