Dgdrxrt

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.

You expanded the brackets, but did not actually group:

$$ 1 + (1+r)s + (1+r+r^2)s^2 + dotsb
= 1 + (s+rs colorred + dotsb) + (s^2+rs^2+r^2s^2colorred + dotsb). $$

Here is the way to group:
beginalign*
& 1+(1+r)s+(1+r+r^2)s^2+dotsb \
&= 1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+dotsb \
&= (1+colorreds+colorreds^2+dotsb)+(colorgreenrs+colorgreenrs^2+dotsb)+(colorbluer^2s^2+r^2s^3+dotsb) \
&= frac11-s+fracrs1-s+fracr^2s^21-s+dotsb \
&= frac11-s(1+rs+r^2s^2+dotsb)
endalign*
Can you finish?

Your sum can be written as
$$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
fracr1-rsum_n=0^infty(rs)^n.$$
Can you take it from here?

P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?

If you multiply your series by $r-1$, you 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 $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$

If you regroup by powers of $r$, the coefficient of $r^n$ is $s^n+s^n+1+s^n+2+cdots=s^nover1-s$, hence

$$1+(1+r)s+(1+r+r^2)s^2+cdots=1+rs+r^2s^2+cdotsover1-s=1over(1-s)(1-rs)$$