Combinatorics identity algebraic proof Prove that:
$$\sum _{k=1}^n \:k\binom{n}{k}^2=n\binom {2n-1}{n-1}$$
I tried to prove it using induction:
For n+1:
$$ \begin{align*} \sum_{k=1}^{n+1} \:k\binom{n+1}{k}^2 &=  \sum \:k\left(\binom{n}{k}+\binom{n}{k-1}\right)^2 \:\left[Pascal's\:rule\right] \\
\\
&= \sum \:k\binom{n}{k}^2+\sum \:2k\binom{n}{k}\binom{n}{k-1}+\sum \:k\binom{n}{k-1}^2 \\
\\
&= n\binom{2n-1}{n-1} +\sum \:2k\binom{n}{k}\binom{n}{k-1} + n\binom{2n-1}{n-2} [*]
\end{align*}$$
[*] - Induction hypothesis 
Now I'm stuck with the middle expression. 
Any advice on how to proceed would be appreciated.
Note: I'm looking for an algebraic proof not combinatorial interpretation. 
 A: $\sum k {n \choose k}^2 = \sum k {n \choose k} {n \choose n -k} $
combinatorial interpretation is that you have a group of $n$ men and $n$ women, you want to choose a team leader who has to be a man and $n-1$ team members - no matter what their gender is. your left hand side corresponds to first saying "ok - my group will have $k \geq 1$ men in it so I choose those men in ${n \choose k}$ ways, choose the leader from them in $k$ ways and then choose $n-k$ women to complete the team in ${n \choose n -k}$ ways. The right hand side corresponds to the method - first choose the leader from $n$ men in $n$ ways then choose the rest of the team $(n-1)$ people from a group of $2n - 1$ in ${2n -1 \choose n-1}$ ways
A: You can prove it directly by using the identities
$$k\binom{n}k=n\binom{n-1}{k-1}$$
and
$$\sum_k\binom{m}k\binom{n}{\ell-k}=\binom{m+n}\ell\;;$$
the latter is Vandermonde’s identity. Both have nice combinatorial proofs that have appeared on this site.
$$\begin{align*}
\sum_{k=1}^nk\binom{n}k^2&=\sum_{k=1}^nn\binom{n-1}{k-1}\binom{n}k\\\\
&=n\sum_{k=1}^n\binom{n-1}{k-1}\binom{n}{n-k}\\\\
&=n\sum_{k=0}^{n-1}\binom{n-1}k\binom{n}{n-1-k}\\\\
&=n\binom{2n-1}{n-1}
\end{align*}$$
