Proving $ \frac{(n!)^2}{n(2n)!}\sum_{k=0}^n k {n \choose k} ^2=\frac{1}{2} $ by induction I'm struggling to prove the following by induction for $n\geq 1$
$$
\frac{(n!)^2}{n(2n)!}\sum_{k=0}^n  k {n \choose k} ^2=\frac{1}{2}
$$
The base case works but trying to prove for $n+1$ is proving difficult, with the algebra getting messy. I think I'm missing something obvious. Trying to prove $m+1$...
\begin{align*}
&=\frac{((m+1)!)^2}{(m+1)(2(m+1))!}\sum_{k=0}^{m+1}  k {m+1 \choose k}^2\\\\
&=\frac{(m+1)^2(m!)^2}{(m+1)(2m+2)(2m+1)(2m)!}\left[m+1+\sum_{k=0}^mk{m+1\choose k}^2\right]\\\\
&=\frac{(m!)^2}{2(2m+1)(2m)!}\left[m+1+\sum_{k=0}^mk{m+1\choose k}^2\right]
\end{align*}
At this point, I'm not sure where to go. Some ideas I have tried are using the fact that
\begin{align*}
{m+1\choose k}&=\frac{m+1}{m+1-k}{m\choose k}
\end{align*}
Or
\begin{align*}
{m+1\choose k}&={m\choose k}+{m\choose k-1}
\end{align*}
But the algebra gets to the point where I feel I am going about it the wrong way. Any help would be appreciated.
 A: Two ways to make the argument without induction.
We know $$k\binom nk^2=k\binom nk\binom n{n-k}$$
and:
$$k\binom nk=n\binom{n-1}{k-1}$$
So we have:
$$\begin{align}
\sum k\binom nk ^2&=n\sum_k \binom{n-1}{k-1}\binom n{n-k}\\&=n\binom{2n-1}{n-1}\\&=n\cdot\frac12\binom {2n}{n}\\
&=\frac{n(2n)!}{2(n!)^2}
\end{align}$$
The second equality is a case of the Vandermonde Identity, and the third is easily shown.
Dividing gives your result.

A combinatorial way to think about it is, if you have $n$ men and $n$ women, the value $n\binom{2n}n$ counts the number of ways to pick a committee of $n$ people including a chairperson, while the sum counts the number of such committees with a woman chairperson, which is half of all the committees. (Basically, $k\binom nk^2$ is the number of such committees with $k$ women, and a woman chair.)
This would also give a more general theorem, if you have $n$ men and $m$ women and a committee of size $c$ with one selected as chair you get:
$$\sum_{k=0}^c k\binom mk\binom n{c-k}=\frac{mc}{m+n}\binom{m+n}c=m\binom{m+n-1}{c-1}.$$
