<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] $$k \binom nk^2=\binom nk\cdot k\binom nk$$
For $k\ge1,$ $$k\binom nk=k\cdot\dfrac{n!}{k!\cdot(n-k)!}=n\dfrac{(n-1)!}{(k-1)!\{n-1-(k-1)\}!}=n\binom{n-1}{k-1}$$
Now in the identity $(1+x)^{2n-1}=(x+1)^n(1+x)^{n-1},$  compare coefficients of $x^n$
$$\binom{2n-1}n=\sum_{k=0}^n\binom nk\binom{n-1}{k-1}$$
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