TITLE: Visual Intuition for the Sum of a FINITE Geometric Series
QUESTION [7 upvotes]: I'm interested in intuitive visual explanations for the sum of a finite geometric series.

I know there are some pretty "intuitive" explanations out there (including some on this site), but I haven't seen any that provide a visual intuition.
If anyone here knows of any and would share them, I'd greatly appreciate it! 
Thanks!

Finite geometric series.
All the answers thus far have been for the infinite case.
Thanks!

REPLY [5 votes]: If anyone would survive being limited to natural numbers only, I suggest an illustration with the help of trees.
Each of the $n$ terms of a geometric series with a common ratio $r$ corresponds to a level in a (perfect) $r$-ary tree $T_n$. The sum $s$ is given by the number of nodes in $T_n$.
The main insight comes from comparing $T_n$ with $T_{n+1}$, i.e. comparing a series of $n$ and $n+1$ terms:

Multiplying $T_n$ with $r$, we get $r$ replicas of $T_n$ that can be rooted in a dummy root, creating a tree $T_{n+1}$ with $sr + 1$ nodes.

Now, we can cover the top part of $T_{n+1}$ by $T_{n}$ leaving only the leaves of $T_{n+1}$, i.e. $r^{n+1}$ nodes:
$rs - s = r^{n+1} - 1$ (we take care of the annoying root by subtracting $1$)

Another way to remove $s$ nodes from $T_{n+1}$ is to cover one of the replicas we created in step 1). This leaves us with $(r-1)$ replicas that should still consist of $r^{n+1} - 1$ nodes: $s(r-1) = r^{n+1} - 1$

We're not interested in $(r-1)$ replicas, but only in one replica, so we divide by $r-1$ and hope that we were not drawing unary trees: $s=\frac{r^{n+1}-1}{r-1}$

In summary, by considering the next term of the series, you increase your sum by a known amount (step 2), which corresponds to multiplying the original by a known amount (step 3).