<s>[INST] 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?
 [/INST] $$
\begin{align}
\sum_{r=0}^{2n}r\binom{2n}{r}^2
&=\sum_{r=1}^{2n}r\binom{2n}{r}\binom{2n}{2n-r}\tag{1}\\
&=\sum_{r=1}^{2n}2n\binom{2n-1}{r-1}\binom{2n}{2n-r}\tag{2}\\
&=2n\binom{4n-1}{2n-1}\tag{3}\\
&=2n\binom{4n-1}{2n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}$
$(3)$: Vandermonde's Identity
$(4)$: $\binom{n}{k}=\binom{n}{n-k}$

Here is a proof of the Vandermonde Identity step using the binomial identity.
The binomial theorem says
$$
\begin{align}
(1+x)^{2n-1}(1+x)^{2n}
&=\sum_{j=0}^{2n-1}\binom{2n-1}{j}x^j\sum_{k=0}^{2n}\binom{2n}{k}x^k\\
&=\sum_{s=0}^{4n-1}\color{#00A000}{\sum_{r=0}^s\binom{2n-1}{r}\binom{2n}{s-r}x^s}
\end{align}
$$
but it also says
$$
(1+x)^{4n-1}=\sum_{s=0}^{4n-1}\color{#00A000}{\binom{4n-1}{s}x^s}
$$
Equating the coefficients of $x^s$, we get
$$
\sum_{r=0}^s\binom{2n-1}{r}\binom{2n}{s-r}=\binom{4n-1}{s}
$$
Setting $s=2n-1$ and substituting $r\mapsto r-1$ gives
$$
\sum_{r=1}^{2n}\binom{2n-1}{r-1}\binom{2n}{2n-r}=\binom{4n-1}{2n-1}
$$
</s>