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

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