$\sum_{k=0}^{2n}(-1)^k\cdot (k-2n)\cdot \binom{2n}{k}^2$ in terms of $n$ and $A$, is If $\displaystyle \sum_{k=0}^{2n}(-1)^k\cdot \binom{2n}{k}^2 = A$, Then value of $\displaystyle\sum_{k=0}^{2n}(-1)^k\cdot (k-2n)\cdot \binom{2n}{k}^2$ in terms of $n$ and $A$, is
$\bf{My\;Try}::$ Using Binomial Theorem for Positive Integral Index::
$\displaystyle (1+x)^{2n} = \binom{2n}{0}+\binom{2n}{1}\cdot x+\binom{2n}{2}\cdot x^2+............+\binom{2n}{2n-1}\cdot x^{2n-1}+\binom{2n}{2n}\cdot x^{2n}$
$\displaystyle (x-1)^{2n} = \binom{2n}{0}\cdot x^{2n}-\binom{2n}{1}\cdot x^{2n-1}+\binom{2n}{2}\cdot x^{2n-2}+......-\binom{2n}{2n-1}\cdot x^{1}+\binom{2n}{2n}$
Now Coeff. of $x^{2n}$ in $(1-x^2)^{2n}$ is
$\displaystyle  = \binom{2n}{0}^2-\binom{2n}{1}^2+\binom{2n}{2}^2+...........-\binom{2n}{2n-1}^2+\binom{2n}{n}^2$
So which is equal to 
$\displaystyle  = \binom{2n}{0}^2-\binom{2n}{1}^2+\binom{2n}{2}^2+...........-\binom{2n}{2n-1}^2+\binom{2n}{n}^2=(-1)\cdot \binom{2n}{n}$
Now for $\displaystyle\sum_{k=0}^{2n}(-1)^k\cdot (k-2n)\cdot \binom{2n}{k}^2 = \sum_{k=0}^{2n}(-1)^k\cdot k\cdot \binom{2n}{k}^2-2n\cdot \sum_{k=0}^{2n}(-1)^k\cdot \binom{2n}{k}^2$
But I Did not understand How can I solve $\displaystyle \sum_{k=0}^{2n}(-1)^k\cdot k\cdot \binom{2n}{k}^2$ in terms of $n$ and $A$
Help Required,
Thanks 
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$\ds{\sum_{k = 0}^{2n}\pars{-1}^{k}\pars{k - 2n}{2n \choose k}^{2}:\ {\large ?}}$

\begin{align}
&\sum_{k = 0}^{2n}\pars{-1}^{k}\pars{k - 2n}{2n \choose k}^{2}
=
\sum_{k = 0}^{2n}\pars{-1}^{k}\pars{k - 2n}{2n \choose k}
\sum_{\ell = 0}^{2n}{2n \choose \ell}\delta_{\ell,k}
\\[3mm]&=
\sum_{k = 0}^{2n}\pars{-1}^{k}\pars{k - 2n}{2n \choose k}
\sum_{\ell = 0}^{2n}{2n \choose \ell}\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,
{1 \over z^{\ell - k + 1}}
\\[3mm]&=
\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,{1 \over z}
\bracks{%
\pars{z\,\partiald{}{z} - 2n}\sum_{k = 0}^{2n}\pars{-1}^{k}{2n \choose k}z^{k}}
\sum_{\ell = 0}^{2n}{2n \choose \ell}\pars{1 \over z}^{\ell}
\\[3mm]&=
\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,{1 \over z}
\bracks{%
\pars{z\,\partiald{}{z} - 2n}\pars{1 - z}^{2n}}\pars{1 + {1 \over z}}^{2n}
\\[3mm]&=
\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,{1 \over z}
\bracks{%
-2nz\pars{1 - z}^{2n - 1} - 2n\pars{1 - z}^{2n}}
{\pars{1 + z}^{2n} \over z^{2n}}
\\[3mm]&=
-2n\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,{1 \over z^{2n + 1}}
\bracks{z\pars{1 + z}\pars{1 - z^{2}}^{2n - 1} - \pars{1 - z^{2}}^{2n}}
\\[3mm]&=
-2n\int_{\vert{z} = 1^{-}}{\dd z \over 2\pi\ic}\,{1 \over z^{2n + 1}}
\bracks{%
\sum_{\ell = 0}^{2n - 1}{2n - 1 \choose \ell}\pars{-1}^{\ell}
\pars{z^{2\ell + 1} + z^{2\ell + 2}}
-
\sum_{\ell = 0}^{2n}{2n \choose \ell}\pars{-1}^{\ell}z^{2\ell}}
\\[3mm]&=
-2n\bracks{%
{2n - 1 \choose n - 1}\pars{-1}^{n - 1}
-
{2n \choose n}\pars{-1}^{n}}
=
2n\pars{-1}^{n}\bracks{{2n - 1 \choose n - 1} + {2n \choose n}}
\end{align}

$${\large%
\sum_{k = 0}^{2n}\pars{-1}^{k}\pars{k - 2n}{2n \choose k}^{2}
=
2n\pars{-1}^{n}\bracks{{2n - 1 \choose n - 1} + {2n \choose n}}}
$$

Notice that $\ds{{-1 \choose -1} = 1}$.
