Formula for $\sum_{k=1}^n k\binom nk^2$. I know that
$$\binom n1^2+\binom n2^2+\binom n3^2+\binom n4^2+\dots+\binom nn^2=\binom{2n}n.$$

Is there a similar formula
$$\binom n1^2+2\binom n2^2+3\binom n3^2+4\binom n4^2+\dots+n\binom nn^2=\cdots?$$

Thanks in advance.
$$r\cdot\left(\binom nr\right)^2=\binom nr\cdot r\cdot\binom nr$$

Now, $$r\binom nr=r\frac{n!}{(n-r)! r!}=r\frac{(n-1)!\cdot n}{\{(n-1)-(r-1)\}!\ (r-1)!\cdot r}=n\binom{n-1}{r-1}$$

Again, $$\frac{(1+x)^{2n-1}}{x^{n-1}}=(1+x)^n\left(1+\frac1x\right)^{n-1}$$

Now, observe that the coefficient of $x$  in the Right Hand Side is $\displaystyle \sum_{0\le r\le n}\binom nr\binom{n-1}{r-1} $ 

What about the Left Hand Side?
HINT:  

The Left Hand Side : $$\frac{(2n-1)!}{( (n-1)! ) ^2}$$