Prove that $$\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=2^{2n}$$
I was having trouble with this one because all four of the indexes are varying here. So I tried to think of a combinatiorial argument, but wasn't quite able to tie everything together. I was thinking along the lines of imagining $2n$ elements which were separated into $n$ pairs, and then partitioned into two subsets (total $2^n$ subset combinations possible). How should I proceed? (PS no proofs using induction please)
Let $ x\in\mathbb{R} $, such that $ \left|x\right|<\frac{1}{4} $. Using cauchy product theorem, we have :
\begin{aligned}\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\binom{2n}{n}x^{n}}\right)^{2}\\ &=\left(\sum_{n=0}^{+\infty}{\binom{-\frac{1}{2}}{n}\left(-4x\right)^{n}}\right)^{2}\\ &=\left(\frac{1}{\sqrt{1-4x}}\right)^{2}\\ &=\frac{1}{1-4x}\\ \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{4^{n}x^{n}}\end{aligned}
Thus for any $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=4^{n} $$
