Prove $\sum\limits_{r=0}^{2n} r \binom{2n}{r}^2 = 2n \binom{4n-1}{2n}$ I expanded 

$(1+x)^{2n}$ = $\sum\limits_{r=0}^{2n} \binom{2n}{r} x^r $

Differentiating both sides, we get 

$2n(1+x)^{2n-1}$ = $0$ + $\binom{2n}{1}$ + $2\binom{2n}{2}x$ + $3\binom{2n}{3}x^2$ .....

Put $x=1$.

As you see, I'm not able to get the 'squares'.Any idea on how to go about it?
Out of $4n$ people, $2n$ men and $2n$ women, select a committee of $2n$ people and choose a woman to be the chair of the committee. If $r$ women are chosen, there are $\binom{2n}{r} \binom{2n}{2n-r} = \binom{2n}{r}^2$ ways to choose the committee and $r$ ways to choose the chair, so the left hand side gives the number of ways to do so. Alternatively, first choose one of the $2n$ women as chair and then select the remaining members of the committee in $\binom{4n-1}{2n-1} = \binom{4n-1}{2n}$ ways.
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$\ds{\sum_{r = 0}^{2n}r{2n \choose r}^{2} = 2n{4n - 1 \choose 2n}:\ {\large ?}}$


  Hereafter we'll use the identity:
  $$\color{#c00000}{%
{m \choose \ell} = \int_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}\,,\qquad m, \ell \in {\mathbb N}\,,\quad m \geq \ell}
$$


\begin{align}
\color{#00f}{\large\sum_{r = 0}^{2n}r{2n \choose r}^{2}}&amp;
=\sum_{r = 0}^{2n}r{2n \choose r}
\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{r + 1}}\,{\dd z \over 2\pi\ic}
=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\sum_{r = 0}^{2n}{2n \choose r}{r \over z^{r}}\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\lim_{\mu \to 1/z}\bracks{\mu\,\partiald{}{\mu}
\sum_{r = 0}^{2n}{2n \choose r}\mu^{r}}\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\lim_{\mu \to 1/z}\bracks{\mu\,\partiald{\pars{1 + \mu}^{2n}}{\mu}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\bracks{{1 \over z}\,2n\pars{1 + {1 \over z}}^{2n - 1}}\,{\dd z \over 2\pi\ic}
=2n\int_{\verts{z} = 1}{\pars{1 + z}^{4n - 1} \over z^{2n + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&amp;=\color{#00f}{\large 2n{4n - 1 \choose 2n}}
\end{align}