Is there a geometrical reason why for $x\geq 0$ and a natural number $n$, $\frac{x^n-1}{x-1 } = \sum_{i=0}^{n-1} x^i$? Let $x \geq0$ and $x$ is a real number. And suppose $n$ is a natural number. Thus,
$$\frac{x^n-1}{x-1 } = \sum_{i=0}^{n-1} x^i$$
I have proven this with induction but It is some sort of interesting to me that what geometrically reason  behind this. While couple days I tried to find a geometrical intuition behind this; but I couldn't find. Now, I suppose it is time to ask for help: First of All, Is there a geometrical thing behind this, and if there is, what is it?

We have $$\forall x \neq 1, \forall n, \sum_{k=0}^{n}x^{k}=\frac{1-x^{n+1}}{1-x}.$$
When $|x|<1$ we can let $n$ tend to infinity to get $\sum_{k=0}^{\infty}x^{k}=\frac{1}{1-x}.$
This is the geometric series. You don't need induction to prove this, just let $S=\sum_{k=0}^{n}x^{k}$ and consider the difference $S-Sx.$
For a geometrical proof of the finite sum of a geometric series consider the following diagram (from here):

The area of the full unit square is equal to the sum of the areas of the yellow rectangles plus the pink square. Thus we have in general:
$$1=(1-r)\sum_{k=0}^{n}r^{k}+r^{n+1},$$
so $$\sum_{k=0}^{n}r^{k}=\frac{1-r^{n+1}}{1-r}$$ as required.
[In this case above we have $n=5.$ For more examples or for the geometrical proof of the infinite sum see here or here]
