Poof that $\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+...+2^n\binom{n}{n}=x^n\text{ for every }n \in \mathbb{N}$

Question

I probably have to use Newton's binomial, but i can't interpret this expression well because i just started stydying Discrete Mathematics and am struggling with this question.

The objective is to proove that exists a x ∈ N

$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+...+2^n\binom{n}{n}=x^n\text{ for every }n \in \mathbb{N}$

Answer 1

By binomial theorem

$$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+…+2^n\binom{n}{n}=(1+2)^n=3^n$$

Answer 2

A possible expansion of $x^n$ using the binomial theorem for $a=1$ and $b=x-1$ is:

$$\forall x \in \mathbb N^* , \ \binom{n}{0}+(x-1)\binom{n}{1}+(x-1)^2\binom{n}{2}+…+(x-1)^n\binom{n}{n}=(1+x-1)^n=x^n$$

I tried to guess your question