TITLE: Binomial sum gives $4^n$
QUESTION [9 upvotes]: I was looking at this question:Swapping the $i$th largest card between $2$ hands of cards
and WolframAlpha gave me this result.
Why is it so? $$\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n?$$

REPLY [11 votes]: Convolution of the series
$$
\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}{n}x^n
$$
gives
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k}x^n
&=\frac1{\sqrt{1-4x}}\frac1{\sqrt{1-4x}}\\
&=\frac1{1-4x}\\
&=\sum_{n=0}^\infty4^nx^n
\end{align}
$$
Equating coefficients of $x^n$ gives
$$
\sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k}=4^n
$$