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

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?

• Apply binomial theorem : $$\sum_{k = 0}^n \binom{n}{k} a^k = (1 + a)^n$$ for $a = 2 x + x^2$. Commented Oct 21, 2022 at 9:56

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.