Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$

Question

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.

Answer 1

I will use the notation $[x^k]\,f(x)$ for denoting the coefficient of $x^k$ in the Taylor/Laurent expansion of $f(x)$ around the origin. We have: $$ S(n)=\sum_{k=0}^{2n}(-1)^k \binom{2n}{k}^2 = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}\binom{2n}{2n-k}=\sum_{\substack{a,b\geq 0 \\ a+b=2n}}(-1)^a\binom{2n}{a}\binom{2n}{b} $$ and since $$ \sum_{c\geq 0}(-1)^c \binom{2n}{c} x^c = (1-x)^{2n}, \qquad \sum_{d\geq 0}\binom{2n}{d} x^d = (1+x)^{2n} $$ it follows that: $$ S(n) = [x^{2n}] (1-x)^{2n}(1+x)^{2n} = [x^{2n}](1-x^2)^{2n} \stackrel{x^2\mapsto z}{=} [z^n](1-z)^{2n}$$ so $S(n) =\color{red}{ (-1)^n \binom{2n}{n}}$ just follows from the binomial theorem.

Answer 2

Think generating functions. The sum is the $x^n$ coefficient of $$\sum_{k=0}^n\binom nkx^k\sum_{l=0}^n(-1)^l\binom nlx^l.$$