<s>[INST] Proving $\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $ [Corrected question]
I'm struggling at proving the following combinatorical identity:
$$\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $$
I would like to see a combinatorical (logical) solution, or an algebraic solution.
 [/INST] Here is what might be the worst possible solution to this problem. We begin like ccorn does by noting
$$\begin{aligned}
\sum_{k=0}^n k\,\binom{n}{k}^2 &= \sum_{k=0}^n k\,\binom{n}{k}^2\,x^{k-1}\bigg|_{x=1} =
\left(\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n \binom{n}{k}^2\,x^k\right)\bigg|_{x=1}
\end{aligned}.$$
We examine
$$\sum_{k=0}^n \binom{n}{k}^2 x^k.$$
Let $y=x^{1/2}$. The above expression can be rewritten as
$$\int_0^{1 } \left(\sum_{k=0}^n \binom{n}{k}y^k e^{2\pi ikt} \right) \left(\sum_{k=0}^n \binom{n}{k}y^k e^{-2\pi ikt} \right) \ dt.$$
Using the binomial theorem, this becomes 
$$\int_0^{1 } \left(1+y e^{2\pi it}\right)^n \left(1+y e^{-2\pi it} \right)^n \ dt= \int_0^1 (1 + y(e^{2\pi i t }+e^{-2\pi i t}) + x)^n \ dt $$
$$= \int_0^1 (1 + 2y\cos(2\pi t)+x)^n \ dt.$$
We want the derivative of this evaluated at $1$. We can pass the derivative with respect to $x$ inside the integral sign, so our desired expression is 
$$ n\int_0^1 \left(\frac{\cos(2\pi t)}{\sqrt x} +1\right) (2\sqrt x \cos(2\pi t) + x +1)^{n-1} \ dt \bigg|_{x=1}=$$
$$ n\int_0^1 \left({\cos(2\pi t)} +1\right) (2 \cos(2\pi t) + 2)^{n-1} \ dt .$$
$$=2^{n-1}n \int_0^1 \left({\cos(2\pi t)} +1\right)^n \ dt .$$
This becomes $$2^{2n-1}n \int_0^1 (\cos(\pi t) )^{2n} \ dt .$$
Using integration by parts, we have the reduction formula 
$$\int_0^1 (\cos(\pi t) )^{2n} \ dt = \frac{2n-1}{2n} \int_0^1 (\cos(\pi t) )^{2n-2} \ dt.$$
This gives the product
$$2^{2n-1}\cdot n \cdot \frac{2n-1}{2n} \frac{2n-3}{2n-2}  \cdots \frac{1 }{ 2}\cdot 1 =2^n \cdot n \cdot \frac{2n-1}{n}\cdot  \frac{2n-3}{n-1}  \cdots \frac{1 }{ 1}\cdot 1 $$
$$= n \frac{(2n-1)!}{(n-1)!n!}$$
Now note that $$\binom{2n-1}{n-1}=\frac{(2n-1)!}{(n-1)!n!},$$
so we recover 
$$n\binom{2n-1}{n-1}$$
as desired.
</s>