Why is this function generate $a_n = 2^n(n+1)$?

Question

Let $G(x) = \frac{1}{(1-x)^2}$ which generates the sequence $a_n = n+1$
How can one infer that $G(2x) = \frac{1}{(1-2x)^2}$ generates $a_n = 2^n(n+1)$?

Thanks.

Answer 1

We have $$\frac{1}{(1-t)^2}=\sum_{n=0}^\infty (n+1)t^n.$$

Replace $t$ everywhere by $2x$. Since $(2x)^n=2^n x^n$, the coefficient of $x^n$ is $(n+1)2^n$.