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.
 A: 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?
A: Your sum can be written as
$$\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}r^k\right)s^n=\sum_{n=0}^{\infty}\frac{1-r^{n+1}}{1-r}s^n=\frac{1}{1-r}\sum_{n=0}^{\infty}s^n-
\frac{r}{1-r}\sum_{n=0}^{\infty}(rs)^n.$$
Can you take it from here?
P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $r\not=1$. What happens when $r=1$?
A: If you multiply your series by $r-1$, you will get$$(r-1)+(r^2-1)s+(r^3-1)s^2+\cdots,\tag1$$which is the sum of$$r+r^2s+r^3s^2+\cdots$$with$$-1-s-s^2-\cdots$$The sum of the first series is $\frac r{1-rs}$, whereas the sum of the second one is $-\frac1{1-s}$. Therefore,$$(1)=\frac1{r-1}\left(\frac r{1-rs}-\frac1{1-s}\right)=\frac1{(1-s)(1-rs)}.$$
A: If you regroup by powers of $r$, the coefficient of $r^n$ is $s^n+s^{n+1}+s^{n+2}+\cdots={s^n\over1-s}$, hence
$$1+(1+r)s+(1+r+r^2)s^2+\cdots={1+rs+r^2s^2+\cdots\over1-s}={1\over(1-s)(1-rs)}$$
