How to prove that $\sum_{k=0}^n\cos(k\pi)\binom{n}{k}=0$?

Question

$$\sum_{k=0}^n\cos(k\pi)\binom{n}{k}=0$$

I approached this problem having no idea that $\cos(k\pi)$ could be substituted so easily. I tried first to expand the sum to $n$. So I wanted to ask if anyone knows how to expand this kind of sum? Would highly appreciate it. Is it like $1\times\cos(k\pi)$ and expand $(1+\cos(k\pi))^n$?

Answer 1

Note that $\cos k\pi=(-1)^k$, so you are asking for $$\sum_{k=0}^n (-1)^k{n \choose k}$$ $$=\sum_{k=0}^n {n \choose k}(1)^{n-k}(-1)^k$$ $$=(1+-1)^n=0^n=0$$ The removal of the sigma comes from the binomial theorem.

ADDED: As Marc van Leeuwen pointed out in a comment, the identity fails for $n=0$. This is because $0^n$ is not zero for $n=0$. Nice catch, Marc!

Answer 2

The desired sum is the real part of

$$\sum_{k=0}^n e^{ik\pi}{n\choose k}=(1+e^{i\pi})^n=0$$