$\displaystyle \sum_{k=1}^{n}k\binom{n}{k}^2=n\binom{2n-1}{n-1}$

Prove this using a combinatorial arguement.

First off, the right hand side can be argued like so. Suppose you have n republicans and n democrats. If you pick one party, you have n ways to select a leader for a committee. Then you have 2n-1 people left to form the committee from of a size n-1? Is this the right reasoning.

As for the left side, suppose you picked the size of the committee k, you have $\displaystyle \binom{n}{k}$ ways to choose the democrats and again with the republicans.