Generating function from $\frac{bx}{1-ax}$

Question

I know that $\displaystyle\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$ (also generates the sequence of 1's).

It is possible to prove from this that $\displaystyle\frac{bx}{1-ax}=\sum_{n=1}^{\infty} ba^{n-1} x^n$.

my question is precisely how it appears $a^{n-1}$.

Answer 1

I am guessing that you are confused by the fact that the exponent for the $a$ term is one lower. I will try to show why this is true.

\begin{align*} \frac{1}{1 - x} &= 1 + x + x^2 + x^3 + \dots\\ \frac{1}{1 - ax} &= 1 + ax + a^2x^2 + a^3x^3 + \dots\\ bx \cdot \frac{1}{1 - ax} &= bx(1 + ax + a^2x^2 + a^3x^3 + \dots)\\ \Rightarrow \frac{bx}{1 - ax} &= bx + bax^2 + ba^2x^3 + ba^3x^4 + \dots\\ &= \sum_{i = 1}^{\infty} ba^{n - 1}x^n. \end{align*}