<s>[INST] How to find the sum of $1+(1+r)s+(1+r+r^2)s^2+\dots$? I was asked to find the geometric sum of the following:
$$1+(1+r)s+(1+r+r^2)s^2+\dots$$
My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:
$$1+(s+rs+\dots)+(s^2+rs^2+r^2s^2+\dots)$$
The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.
Any help would be appreciated.
 [/INST] You expanded the brackets, but did not actually group:

$$ 1 + (1+r)s + (1+r+r^2)s^2 + \dotsb
   = 1 + (s+rs \color{red}{{} + \dotsb}) + (s^2+rs^2+r^2s^2\color{red}{{} + \dotsb}). $$

Here is the way to group:
\begin{align*}
& 1+(1+r)s+(1+r+r^2)s^2+\dotsb \\
&= 1+(\color{red}{s}+\color{green}{rs})+(\color{red}{s^2}+\color{green}{rs^2}+\color{blue}{r^2s^2})+\dotsb \\
&= (1+\color{red}{s}+\color{red}{s^2}+\dotsb)+(\color{green}{rs}+\color{green}{rs^2}+\dotsb)+(\color{blue}{r^2s^2}+r^2s^3+\dotsb) \\
&= \frac{1}{1-s}+\frac{rs}{1-s}+\frac{r^2s^2}{1-s}+\dotsb \\
&= \frac{1}{1-s}(1+rs+r^2s^2+\dotsb)
\end{align*}
Can you finish?
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