Alternating sum involving binomial coefficient

Question

How do I find the following sum? $\sum_{k=0}^n(-1)^k{2n\choose2k}$

Tried to simplify it somehow but got nothing less complicated.

Answer 1

Hint:

\begin{align} \sum\limits_{k=0}^{n} {{2n}\choose{2k}} (-1)^k &= \text{ even terms of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\ &= \text{ real part of } \sum\limits_{k=0}^{2n} {{2n}\choose{k}} i^k \\ &= \text{ real part of } (1+i)^{2n} \end{align}

Answer 2

Hint. One may observe that, by the binomial theorem, $$ \begin{align} \sum_{k=0}^n(-1)^k{2n\choose2k}&=\sum_{k=0}^n {2n\choose2k}i^{2k} \\\\&=\sum_{k=0}^{2n} {2n\choose2k}i^{2k} \\\\&=\sum_{p=0}^{2n} {2n\choose p}\frac{i^{p}+(-i)^p}2 \\\\&=\dfrac{(1+i)^{2n}+(1-i)^{2n}}{2} \\\\&=(2i)^{n}\frac{1+(-1)^n}2. \end{align} $$