TITLE: Summation simplification $\sum_{k=0}^{n} \binom{2n}{k}^2$
QUESTION [12 upvotes]: $\sum_{k=0}^{n} \binom{2n}{k}^2$
So i'm trying to simplify this one and I'm stuck in nowhere. Some kind of tip would be appreciated.
Thanks! :)

REPLY [2 votes]: Using the identities
$$
\sum_{k=0}^p\binom{n}{k}\binom{m}{p-k}=\binom{n+m}{p}
$$
and
$$
\binom{n}{n-k}=\binom{n}{k}
$$
and mapping $n,m,p\mapsto2n$, we get
$$
\begin{align}
\binom{4n}{2n}
&=\sum_{k=0}^{2n}\binom{2n}{k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=n+1}^{2n}\binom{2n}{k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=0}^{n-1}\binom{2n}{2n-k}^2\\
&=\sum_{k=0}^n\binom{2n}{k}^2+\sum_{k=0}^{n-1}\binom{2n}{k}^2\\
&=2\sum_{k=0}^n\binom{2n}{k}^2-\binom{2n}{n}^2\\
\sum_{k=0}^n\binom{2n}{k}^2&=\frac12\binom{4n}{2n}+\frac12\binom{2n}{n}^2
\end{align}
$$