<s>[INST] Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I haven't been able to make much progress in solving it.
My approach:
\begin{align}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{2} = \binom{2n}{0}^{2} - \binom{2n}{1}^{2} + \binom{2n}{2}^{2} - ... -\binom{2n}{2n-1}^{2} + \binom{2n}{2n}^{2} \\
= \binom{2n}{0}.\binom{2n}{0} - \binom{2n}{1}.\binom{2n}{1} + \binom{2n}{2}.\binom{2n}{2} - ... -\binom{2n}{2n-1}.\binom{2n}{2n-1} + \binom{2n}{2n}.\binom{2n}{2n} \\
\text{By Symmetry of binomial coefficients} \\
= \binom{2n}{2n}.\binom{2n}{0} - \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} - ... -\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} \\
= \binom{2n}{2n}.\binom{2n}{0} + \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} + ... +\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} - 2.\left(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}\right) \\
\text{By Vandermond's identity, first component, i.e.not enclosed within -2.(...) evaluates to C(4n, 2n)} \\
\binom{4n}{2n} - 2.(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1})
\end{align}
I'm lost beyond this point. It will be extremely helpful if someone can direct me in the right direction or provide the answer to this perplexing and challenging problem. Thank you.
 [/INST] Think generating functions. The sum is the $x^n$ coefficient of
$$\sum_{k=0}^n\binom nkx^k\sum_{l=0}^n(-1)^l\binom nlx^l.$$
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