Show that for every even $n$, ${^nC}_0 - 2\cdot{^nC}_1 + 2^2\cdot{^nC}_2 -\dotsb+2^n\cdot{^nC}_n = 1$

Question

Solve by Binomial Expansion.

Show that for every even $n$ $${^nC}_0 - 2\cdot{^nC}_1 + 2^2\cdot{^nC}_2 -\dotsb+2^n\cdot{^nC}_n = 1$$

$(1+(-2))^n = {^nC}_0-{^nC}_1\cdot 2+{^nC}_2\cdot 2^2+\dotsb+2^n\cdot {^nC}_n$

$(-1)^n$ is even only if $n$ is even,
Hence, $1=(1+(-2))^n $
And if $n$ is even $2^n$ is also even. Hence proved. Is this correct?

Answer 1

$1 = 1^n = (1+(-2))^n $ as $2$ is the given at the end i.e \begin{align*} 2^n &= 2^0\binom{n}{0}-2^1\binom{n}{n-1}+2^2\binom{n}{n-2}+\dotsb+ 2^n\binom{n}{n-n} \\ &= 2^0 - 2^1\binom{n}{1}+ 2^2\binom{n}{2}+\dotsb+ 2^n \end{align*}