Sum of square of binomial coeffcient with positive and negative terms Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$
What I've tried:
$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$
$$-2n(1-x)^{2n-1}=-\binom{2n}{1}+2\binom{2n}{2}x-3\binom{2n}{3}x^2+\cdots +n\binom{2n}{2n}x^{2n-1}$$

Sum notation:
$$\sum_{k=0}^{2n} (-1)^{k-1}k\binom{2n}{k}^2$$

 A: 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 $x=0$: 
$S(0)= 2n2^{2n-1}P_{2n-2}(0)\tag7$
A: Starting from
$$\sum_{k=0}^{2n} (-1)^{k-1} k {2n\choose k}^2
= \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
k {2n\choose k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose k-1}
= 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose 2n-k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
[z^{2n-k}] (1+z)^{2n-1}
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\left(1+\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k\right).$$
Now $2n [z^{2n}] (1+z)^{2n-1}$ is zero, so we may continue with
$$2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = - 2n [z^{2n}] (1+z)^{2n-1} (1-z)^{2n}
= - 2n [z^{2n}] (1-z^2)^{2n-1} (1-z)
\\ = - 2n [z^{2n}] (1-z^2)^{2n-1}
= - 2n [z^{n}] (1-z)^{2n-1}.$$
This is
$$(-1)^{n+1} \times 2n \times {2n-1\choose n}
= (-1)^{n+1} \times 2n \times {2n\choose n}
\frac{n}{2n}$$
for an answer of
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{n+1} \times n \times {2n\choose n}.}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{2n}\pars{-1}^{k - 1}\, k\,
{2n \choose k}^{\!\! 2}} =
\left.\partiald{}{x}\sum_{k = 0}^{2n}x^{k}{2n \choose k}
{2n \choose 2n - k}\,\right\vert_{\ x\ =\ -1}
\\[5mm] = &\
\left.\partiald{}{x}\sum_{k = 0}^{2n}x^{k}{2n \choose k}
\bracks{z^{2n - k}}\pars{1 + z}^{2n}\,\right\vert_{\ x\ =\ -1}
\\[5mm] = &\
\left.\bracks{z^{2n}}\pars{1 + z}^{2n}\,
\partiald{}{x}\sum_{k = 0}^{2n}{2n \choose k}\pars{zx}^{k}
\,\right\vert_{\ x\ =\ -1}
\\[5mm] = &\
\left.\bracks{z^{2n}}\pars{1 + z}^{2n}\,
\partiald{\pars{1 + zx}^{2n}}{x}\,\right\vert_{\ x\ =\ -1}
\\[5mm] = &\
\bracks{z^{2n}}\pars{1 + z}^{2n}\,
\bracks{2n\pars{1 - z}^{2n -1}\, z}
\\[5mm] = &\
2n\bracks{z^{2n - 1}}\bracks{%
\pars{1 + z}^{2n - 1} + z\pars{1 + z}^{2n - 1}}\pars{1 - z}^{2n - 1}
\\[5mm] = &\
2n\braces{\underbrace{\bracks{z^{2n - 1}}\pars{1 - z^{2}}^{2n - 1}}
_{\ds{\ =\ 0}}\ +\
\bracks{z^{2n - 2}}\pars{1 - z^{2}}^{2n - 1}}
\\[5mm] = &\
2n\bracks{{2n - 1 \choose n - 1}\pars{-1}^{n - 1}} =
\bbx{\large\pars{-1}^{n - 1}\, n^{2}{2n \choose n}} \\ &
\end{align}
