What is the $n$-th sequence element for the generating function $\frac{1}{(1-ax)^2}$?

Question

for e.g. for $\frac{1}{(1-ax)} = a^n$ or for $\frac{1}{(1-x)^2} = n+1$

generating function = $\frac{1}{(1-ax)^2}$

Answer 1

Method 1:

Use the general formula for the $k$-th derivative of a power (or establish through induction):

$$\frac{\mathrm d^k}{\mathrm dx^k}\frac1{(1-ax)^2}=\frac{(k+1)!a^k}{(1-ax)^{k+2}}$$

Method 2:

Since $\dfrac1{(1-ax)^2}=\dfrac1{(1-ax)}\cdot\dfrac1{(1-ax)}$, and $\dfrac1{(1-ax)}$ is the generating function for $a^k$, the coefficients of $\dfrac1{(1-ax)^2}$ are given by the autoconvolution of $a^k$; i.e. the sum

$$\sum_{j=0}^k a^j a^{k-j}$$

whose simplification I leave to you.

Answer 2

You give that $\frac{1}{(1-x)^2}$ is the generating function of $a_n=n+1$. That means that $$ \frac{1}{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k\tag{1} $$ Therefore, $$ \frac{1}{(1-ax)^2}=\sum_{k=0}^\infty (k+1)(ax)^k\tag{2} $$ Thus, $\frac{1}{(1-ax)^2}$ is the generating function for $a_n=(k+1)a^k.$