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.
 A: $$r\cdot\left(\binom nr\right)^2=\binom nr\cdot r\cdot\binom nr$$
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?
A: I accidentally stumbled across this question and have another solution -
Consider 
$$
(1 + x)^n \times (1+x)^n = \Bigg(\binom {n} {0} x^0 + \binom {n} {1} x^1 + \binom {n} {2} x^2 + \cdots + \binom {n} {n} x^n \Bigg) ^ 2 = (1+x)^{2n} = \Bigg(\binom {2n} {0} x^0 + \binom {2n} {1} x^1 + \binom {2n} {2} x^2 + \cdots + \binom {2n} {2n} x^{2n} \Bigg)
$$
Now take derivative of both sides of the binomonial expansions to obtain  -
$$
2 \times \Bigg(\binom {n} {0} x^0 + \binom {n} {1} x^1 + \binom {n} {2} x^2 + \cdots+ \binom {n} {n} x^n \Bigg) \times \Bigg(1\binom {n} {1} x^0 + 2\binom {n} {2} x^1 + \cdots + n\binom {n} {n} x^{n-1} \Bigg) = \Bigg(1\binom {2n} {1} x^0 + 2\binom {2n} {2} x^1 + \cdots + 2n\binom {2n} {2n} x^{2n-1} \Bigg)
$$
In this identity equate coefficients of $$x^{n-1}$$ to obtain -
$$
2 \times \Bigg(1 \binom {n} {n -1} \binom {n} {1} x^0 x^{n-1} + 2 \binom {n} {n-2} \binom {n} {2} x^1 x^{n-2} + \cdots + n\binom {n} {n} \binom {n} {0} x^{n-1} x^{0} \Bigg) = n \binom {2n} {n} x ^ {n-1}
$$
Applying the identity $$\binom {n} {a} = \binom {n} {n-a}$$ we have -
$$
1 {\binom {n} {1}} ^ 2 + 2 {\binom {n} {2}} ^ 2 + 3 {\binom {n} {3}} ^ 2 + \cdots + n {\binom {n} {n}} ^ 2 = \frac {n} {2} \binom {2n} {n}
$$
A: At first.
$$
\binom n1^2+\binom n2^2+\binom n3^2+\binom n4^2+\dots+\binom nn^2=\binom{2n}n-1.
$$
Second, notice that.
$$
1\binom n1^2+2\binom n2^2+3\binom n3^2+4\binom n4^2+\dots+(n-1)\binom n{n-1}^2=\\(n-1)\binom n1^2+(n-2)\binom n2^2+(n-3)\binom n3^2+(n-4)\binom n4^2+\dots+1\binom n{n-1}^2.
$$
A: HINT: 
The Left Hand Side : $$\frac{(2n-1)!}{( (n-1)! ) ^2}$$
