Prove that $\sum_{k=1}^nk^2{n\choose k}^2=n^2 \binom {2n-2}{n-1}$ Please help me / give a hand with combinational prove for:
$$ 1^2 \binom  n 1 ^2 + 2^2 \binom n 2 ^2 + \dots + n^2 \binom n n ^2 = n^2 \binom {2n-2}{n-1}$$

By  way of  enrichment here  is another  solution using  basic complex
variables.

Start by restating the problem using the alternate binomial coefficient:
we seek to show that
$$\sum_{k=0}^n {n\choose k} k^2 {n\choose n-k}
= n^2 {2n-2\choose n-1}.$$
Introduce the integral representation
$${n\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-k+1}} \; dz.$$
This gives the following integral for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}
\sum_{k=0}^n {n\choose k} k^2 z^k \; dz.$$
Now recall that
$$(1+z)^n = \sum_{k=0}^n {n\choose k} z^k$$
and hence
$$nz (1+z)^{n-1} = \sum_{k=0}^n {n\choose k} k z^k$$
and finally
$$z(n (1+z)^{n-1} + nz(n-1)(1+z)^{n-2})
= \sum_{k=0}^n {n\choose k} k^2 z^k.$$
Substituting this into the sum gives
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\left(n \frac{(1+z)^{2n-1}}{z^n}
+ n(n-1) \frac{(1+z)^{2n-2}}{z^{n-1}} \right)\; dz.$$
Extracting coefficients we finally obtain
$$n {2n-1\choose n-1} + n(n-1) {2n-2\choose n-2}
\\= n \frac{2n-1}{n} {2n-2\choose n-1} 
+ n(n-1) \frac{n-1}{n} {2n-2\choose n-1}
\\ = (2n-1 + (n-1)^2) {2n-2\choose n-1}
= n^2 {2n-2\choose n-1}.$$

Apparently this method is due to Egorychev.
A trace as to when it appeared on MSE and by whom starts at this
MSE link.
