Visual Intuition for the Sum of a FINITE Geometric Series 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!

The triangle/trapezoid method is probably the most conventional, however there is another method constructed by simplifying it.
We start by expressing each term in $r^0 + r^1 + \dots  + r^n$ as a rectangle of area $r^i$ and side length $r^i - r^{i + 1}$:

Notice the other side length must be: $\frac{1}{1-r}$, since $r^i / (r^i - r^{i + 1}) = \frac{1}{1 - r}$
Now, joining these rectangles from $0$ to $n$, we find the side lengths add up to: $r^0 - r^{n+1}$
$$(r^0 - r^1) + (r^1 - r^2) + \dots + (r^{n-1} - r^n) + (r^n - r^{n+1}) = r^0 - r^{n + 1}$$

It tells us the rectangle with side lengths $r^0 - r^{n + 1}$ and  $\frac{1}{1-r}$ has area: $r^0 + r^1 + \dots + r^n$
So multiplying the side lengths produces the desired expression:
$$(r^0 - r^{n + 1}) \cdot \left( \frac{1}{1-r} \right) = \frac{r^0 - r^{n+1}}{1-r} = r^0 + r^1 + \dots +r^n$$
Commonly we evaluate $r^0 = 1$ and an arbitrary $a$ is added for scaling:
$$\frac{a(1 - r^{n+1})}{1-r} = ar^0 + ar^1 + \dots +ar^n$$
I imagine someone has come up with this exact method before, however I was not able find a similar method anywhere online.
