<s>[INST] Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$ 
Prove that
  $$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$

So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum_{k=0}^{n-1} \frac{(n-1)!n!}{(n-k-1)!k!(n-k-1)!(k+1)!}
= n^2 \sum_{k=0}^{n-1} \frac{(n-1)!^2}{(n-k-1)!^2k!^2(k+1)}
=n^2 \sum_{k=0}^{n-1} \binom{n-1}{k}^2\frac{1}{k+1}.
$$
I do not know what to do with $\frac{1}{k+1}$, how to get rid of that.
 [/INST] We  present a  slight  variation  using formal  power  series and  the
coefficient-of operator. Starting from
$$\sum_{k=1}^n k {n\choose k}^2
= \sum_{k=1}^n k {n\choose k} [z^{n-k}] (1+z)^n
\\ =  [z^n] (1+z)^n \sum_{k=1}^n k {n\choose k} z^k
=  n [z^n] (1+z)^n \sum_{k=1}^n  {n-1\choose k-1} z^k
\\ =  n [z^n] z (1+z)^n \sum_{k=0}^{n-1}  {n-1\choose k} z^k
=  n [z^{n-1}] (1+z)^n (1+z)^{n-1}
\\ = n [z^{n-1}] (1+z)^{2n-1} = n \times {2n-1\choose n-1}$$
which is the claim.
</s>