$\sum_{m = 0}^n\binom{n}{2m}(-1)^m = 2^{\frac{n}{2}}\cos\left(\frac{n\pi}{4}\right)$

Question

How can I prove that $S_1(n)=\sum_{m = 0}^n\binom{n}{2m}(-1)^m$ is equal to $2^{\frac{n}{2}}\cos\left(\frac{n\pi}{4}\right)$ using the binomial expansion $(1+z)^n=\sum_{r=0}^n\binom{n}{r}z^r$?

score 1 · Accepted Answer · answered Mar 21 '21 at 17:22

1

Hint

As $-1=i^2,$ let $z=i$

Use $$(1+z)^n+(1-z)^n=?$$

and $1+i=\sqrt2e^{i\pi/4}$ using Intuitive explanation of Euler's formula $e^{it}=\cos(t)+i\sin(t)$

answered Mar 21 '21 at 17:22

lab bhattacharjee

274,582

$\sum_{m = 0}^n\binom{n}{2m}(-1)^m = 2^{\frac{n}{2}}\cos\left(\frac{n\pi}{4}\right)$

1 Answers1