# Alternating sum involving binomial coefficient

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.

• Maybe the binomial theorem, $(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$, could help. Maybe with $2n$ instead of $n$, and $x=1$ and $y=-1$ ? Then see if you can manipulate it into what you actually want. – Pixel Oct 27 '16 at 14:56
• Look closely at $(1+i)^{2n}$ and $(1-i)^{2n}$. – Daniel Fischer Oct 27 '16 at 14:56

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}
• Why is your last line not $\dfrac{(1+i)^{2n}+(1-i)^{2n}}{2}$? – Henry Oct 27 '16 at 15:03
• @Henry Because $(1+i)^{2n}=\big(\sqrt{2}e^{i\pi/4}\big)^{2n}=(2i)^n$. – Olivier Oloa Oct 27 '16 at 15:27
• OK. So you are saying the sum is $2^n$ when $n$ is a multiple of $4$, it is $-2^n$ when $n$ is even but not a multiple of $4$, and is $0$ when $n$ is odd. That works for me. – Henry Oct 27 '16 at 15:39