close form for $\sum_{k=0}^n\binom{2n}{n} \binom{2n-2k}{n-k}$

Question

Is it possible to find a closed form for the expression below?

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

I think it is possible to use a former solved one below.

$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom{2n}{n} x^n $

Answer 1

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 : $$ \sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=4^{n} $$