TITLE: Summation of weighted squares of binomial coefficients
QUESTION [6 upvotes]: Show that 
$$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$

REPLY [10 votes]: $$
\begin{align}
\sum_{k=0}^n\left[\frac{n-2k}{n}\binom{n}{k}\right]^2
&=\sum_{k=0}^n\left[\binom{n}{k}-2\binom{n-1}{k-1}\right]^2\\
&=\sum_{k=0}^n\left[\binom{n-1}{k}-\binom{n-1}{k-1}\right]^2\\
&=\sum_{k=0}^n\binom{n-1}{k}^2+\binom{n-1}{k-1}^2-2\binom{n-1}{k}\binom{n-1}{k-1}\\
&=2\binom{2n-2}{n-1}-2\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-k}\\
&=2\binom{2n-2}{n-1}-2\binom{2n-2}{n}\\
&=2\binom{2n-2}{n-1}-2\frac{n-1}{n}\binom{2n-2}{n-1}\\
&=\frac2n\binom{2n-2}{n-1}
\end{align}
$$