Show with the help of binomial theorem that these two equations are equal?

Question

Show with the help of binomial theorem that these two expression are equal for $n\ge 0$ then this $$ \sum_{k=0}^n \binom n k x^k (2+x)^k = \sum_{k=0}^{2n} \binom {2n} k x^k $$

I don’t know how to do it but here is the answer.

\begin{align} \sum_{k=0}^n \binom n k x^k (2+x)^k &= \sum_{k=0}^n \binom n k (2x+x^2)^k \\ &= [\text{binomi} \\ &= (1+2x+x^2)^n \\ &= (1+x)^{2n} \\ &= \sum_{k=0}^{2n} \binom {2n} k x^k. \end{align}

I don’t understand why they were able to multiply $x^k(2+x)^k$ and then get $(1+2x+x^2)$. How did binomial theorem make it possible?

Answer 1

First step is just $a^n b^n = (ab)^n$.

Second step is: \begin{align} \sum_{k=0}^n \binom n k (2x+x^2)^k &= \sum_{k=0}^n \binom n k \color{blue}{\boldsymbol{1^{n-k}}} (2x+x^2)^k \\ &= \text{Binomial expansion} \\ &\qquad\qquad \text{of $1$ and $2x+x^2$} \\ &= \big(1+2x+x^2\big)^n \\ &= \big((1+x)^2\big)^n \\ &= (1+x)^{2n}. \end{align} Then apply the binomial theorem.