Prove $\sum_{k=0}^n\binom{2n+1}{2k}=4^n$

Question

I once had to show that $\cos(x)\sin(x)=\frac{1}{2}\sin(2x)$ using the Cauchy product and relied on $$\sum_{k=0}^n\binom{2n+1}{2k}=4^n.$$ However I never came up with a proof why this is true - is there any short proof why this holds?

Answer 1

$2^{2n+1} =(1+1)^{2n+1} =\sum_{k=0}^{2n+1} \binom{2n+1}{k} $.

$0 =(1-1)^{2n+1} =\sum_{k=0}^{2n+1} (-1)^k\binom{2n+1}{k} $.

Adding these, $2^{2n+1} =\sum_{k=0}^{2n+1} \binom{2n+1}{k}(1+(-1)^k) =2\sum_{k=0}^{n} \binom{2n+1}{2k} $ so $2^{2n} =\sum_{k=0}^{n} \binom{2n+1}{2k} $.