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I've been wondering how to work out the compact form of the following. $$\sum^{50}_{k=1}\binom{101}{2k+1}^{2}$$

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$$\begin{align}\sum_{k=0}^m \binom {2m+1}{2k+1}^2 &=\sum_{k=0}^m \binom {2m+1}{2k+1}\binom {2m+1}{2m-2k} \color{lightgrey}{=\sum_{j=0}^m\binom {2m+1}{2(m-j)+1}\binom {2m+1}{2j}\quad \scriptsize (j=m-k)}\\ &=\frac 12 \sum_{k=0}^m \binom {2m+1}{2k}\binom {2m+1}{2(m-k)+1}+\binom {2m+1}{2k+1}\binom{2m+1}{2m-2k}\\ &=\frac 12 \sum_{i=0}^{2m+1}\binom {2m+1}i\binom {2m+1}{2m+1-i}\\ &=\frac 12 \binom {4m+2}{2m+1}\\ \sum_{k=1}^m \binom {2m+1}{2k+1}^2&=\frac 12 \binom {4m+2}{2m+1}-\binom {2m+1}1^2\\ &=\frac 12 \binom {4m+2}{2m+1}-(2m+1)^2 \end{align}$$ Put $m=50$: $$\sum_{k=1}^{50}\binom {101}{2k+1}^2=\color{red}{\frac 12 \binom {202}{101}-101^2}\qquad\blacksquare$$

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    $\begingroup$ You are correct! $\endgroup$
    – user940
    Jun 15, 2016 at 16:25

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