How do I prove this statement involving binomial coefficients?

Question

Prove for an even $n$,

$$ \binom{n}{0}-2 \binom{n}{1} + 2^2 \binom{n}{2} - \cdots + 2^n \binom{n}{n} = 1$$

I know induction but I don't think it will work since it's only true for even $n$? Substituting an odd $n$, the result comes out to be $-1$

P.S – this is a homework problem. I'm meant to do it on my own. So I'd really appreciate hints. Please note that this is high school level problem, so I don't know of any badass proving methods like y'all. Just couple of basics like induction and contradiction. Thanks!

Answer 1

Given n is even.

$$(1-x)^n=\binom{n}{0}-x\binom{n}{1}+x^2\binom{n}{2}-x^3\binom{n}{3}+\cdots +x^n\binom{n}{n}$$

put $$x=2$$ $$(1-2)^n=\binom{n}{0}-2\binom{n}{1}+2^2\binom{n}{2}-2^3\binom{n}{3}+\cdots +2^n\binom{n}{n}$$ $$\binom{n}{0}-2\binom{n}{1}+2^2\binom{n}{2}-2^3\binom{n}{3}+\cdots +2^n\binom{n}{n}=(-1)^n=1$$

since $n$ is even $(-1)^n=1$

Answer 2

$$\left(1-x\right)^n= \sum_{k=0}^n \binom{n}{k}(-x)^k$$

With $x=2$ and $n$ even.