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.

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 -
$$
1 {\binom {n} {1}} ^ 2 + 2 {\binom {n} {2}} ^ 2 + 3 {\binom {n} {3}} ^ 2 + \cdots + n {\binom {n} {n}} ^ 2 = \frac {n} {2} \binom {2n} {n}
$$
