Summation simplification $\sum_{k=0}^{n} \binom{2n}{k}^2$

Try this: $$\sum_{k = 0}^n \binom{2n}{k}^2 = \sum_{k = 0}^n \binom{2n}{k} \binom{2n}{2n - k} = \frac{1}{2} \sum_{k = 0}^{2n} \binom{2n}{k} \binom{2n}{2n - k} + \frac{1}{2}\binom{2n}{n}^2.$$ Then the sum turns into the coefficient described by anon in the answer below: the $x^{2n}$ term in the product $$(1 + x)^{2n} (1 + x)^{2n} = (1 + x)^{4n}$$ which is $\binom{4n}{2n}$. So the answer is $$\frac{1}{2}\binom{4n}{2n} + \frac{1}{2} \binom{2n}{n}^2.$$

It suffices to find the summation indexed from $k=0$ to $k=2n$. This will be the constant coefficient of $(1+x)^{2n}(1+x^{-1})^{2n}=(1+x)^{4n}x^{-2n}$, or the $x^{2n}$ coefficient of $(1+x)^{4n}$.

Count the cardinality of $\mathcal A=\left\{(A,B):A\subseteq\left\{1,2,\ldots,2n\right\},B\subseteq\left\{2n+1,2n+2,\ldots,4n\right\},|A|=|B|\right\}$ in two ways.

Way 1: For each $k \in \left\{0,1,\ldots 2n\right\}$ choose $k$ elements for $A$ and $k$ elements for $B$. $$\displaystyle|\mathcal A|=\sum_{k=0}^{2n}{2n\choose k}{2n\choose k}=2\sum_{k=0}^{n}{2n\choose k}{2n\choose k}-{2n\choose n}^2$$

Way 2: For any $S\subseteq\left\{1,2,\ldots,4n\right\}$ with $|S|=2n$ take $A=S\cap\left\{1,2,\ldots,2n\right\}$ and $B=S^c\cap\left\{2n+1,2n+2,\ldots,4n\right\}$ where $S^c=\left\{1,2,\ldots,4n\right\}\setminus S$. Therefore $$\left|\mathcal A\right| = \left|\left\{S\subseteq\left\{1,2,\ldots,4n\right\}:|S|=2n\right\}\right|={4n\choose 2n}$$