I was checking some theoretical properties with a concrete example. At some point, I needed to use Wolfram Alfa in order to get the equality
$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k} = 4^k .$$
So now, I have two reasons to know this equality is true: A blind trust in Wolfram, and the fact that this equality must be true for all $k\in \mathbb{N}$ as an indirect byproduct of an unrelated proposition.
This is a little unsatisfactory, and I wanted to find some direct proof, bonus points if it use combinatorics, but I couldn't work it out or find an answer online.
Maybe some of you know some tricks that make this equality obvious?
