My approach: change $4^n$ to $2^{2n}$, then we're trying to show $\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=\sum_{k=0}^{2n} \binom{2n}{k}$, but then I got stuck. Any help would be appreciated.
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See e.g. Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ or Proving $\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n$. – Minus One-Twelfth Jun 05 '19 at 00:22
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1Oh. Sorry for the duplicate. I couldn't find another one. – Kai Jun 05 '19 at 00:22