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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? $$\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n?$$

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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 $$ \sum_{k=0}^n\binom{2k}{k}\binom{2n-2k}{n-k}=4^n $$

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