Sum of square of binomial coeffcient with positive and negative terms

The solution uses the series representation of Legendre polynomials:

$P_n(x)=\frac{1}{2^n}\sum\limits_{k=0}^n \binom{n}{k}^2(x-1)^{n-k}(x+1)^k\tag1$

$\frac{x+1}{x-1}=-1$ at $x=0$ is valid. Extend the original sum (S) in the following way:

$S\frac{(x-1)^{2n}}{2^{2n}}=\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 k(\frac{x+1}{x-1})^{k-1}(x-1)^{2n}\tag2$

We can realize that

$k(\frac{x+1}{x-1})^{k-1}=(-\frac{(x-1)^2}{2})\frac {d}{dx}(\frac{x+1}{x-1})^{k}\tag3$

Put it back to eqution (2) and replace the order of sum and derivation we get:

$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\frac {d}{dx}\Big(\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 (\frac{x+1}{x-1})^{k}(x-1)^{2n}\Big)\tag4$

Let's compare the summa part of (4) and $P_n(x)$ we get:

$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\dfrac {dP_{2n}(x)}{dx}\tag5$

Finally

$S(x)=\frac{2^{2n-1}}{(x-1)^{2n-2}}\dfrac {dP_{2n}(x)}{dx}\tag6$

Applying the recursion relation of the Legendre polynomials we have at $x=0$:

$S(0)= 2n2^{2n-1}P_{2n-2}(0)\tag7$