Why does the geometric series formula intuitively work? I understand the proof for the geometric series formula, but I don't understand how the formula, $S_n = a_1\frac{(r^n -1)}{(r-1)}$
actually relates to the sum of all the terms. 
What operations are taking place in the formula to give the sum.
 A: I like to think of it for integers working over different bases.
For example, $9999 = 10000 -1$ is obvious and intuitive.  Written another way, this reads:
$$
10^0 + 10^1 + \cdots + 10^3 = \frac{10^4 - 1}{10 -1}
$$
Then just replace the $10$ with $x$.
A: Try subtracting these two,
$$\begin{align}S&=1+r+r^2+\dots+r^n\\-(\ rS&=\quad\ \ \ r+r^2+\dots+r^n+r^{n+1}\ )\\\hline(1-r)S&=1+0+0\phantom{^2}+\dots+0^\phantom{^n}-r^{n+1}\end{align}$$
So, we have
$$(1-r)S=1-r^{n+1}$$
$$S=\frac{1-r^{n+1}}{1-r}$$
So that we now have
$$1+r+r^2+\dots+r^n=\frac{1-r^{n+1}}{1-r}$$
A: We have a geometric sequence (or progression), in our case finite with $\;n\;$ terms:
$$a_1,\,a_1r,\,a_1r^2,\,\ldots,\,a_1r^{n-1}$$
Let's call the sum of the above sequence $\;S\;$ , so
$$\begin{align*}&S=\sum_{k=0}^{n-1}a_1r^k=a_1\left(1+r+\ldots+r^{n-1}\right)\implies\\{}\\&rS=a_1\left(r+r^2+\ldots+r^n\right)\implies\\{}\\&(1-r)S=a_1\left(1+r+\ldots+r^{n-1}-r-r^2-\ldots-r^{n-1}-r^n\right)=a_1(1-r^n)\implies\\{}\\&S=a_1\frac{1-r^n}{1-r}\end{align*}$$
Observe that in the above there exists hidden induction.
A: If ever in doubt, you can try induction.  Notice that
$$a_1+a_1r=a_1\frac{1-r^2}{1-r}$$
So the formula is true for $n=1$.
Suppose that $a_1+a_1r+a_1r^2+\dots+a_1r^k=a_1\frac{1-r^{k+1}}{1-r}$ for some $k$.  See then that
$$S=a_1+a_1r+a_1r^2+\dots+a_1r^k+a_1r^{k+1}\\=a_1\frac{1-r^{k+1}}{1-r}+a_1r^{k+1}\\=a_1\frac{1-r^{k+2}}{1-r}$$
This tells us that if the formula is true for $k=1:$
$$a_1+a_1r=a_1\frac{1-r^2}{1-r}$$
Then it is also true for $k+1:$
$$a_1+a_1r+a_1r^2=a_1\frac{1-r^3}{1-r}$$
And that for any $p=k+1$, it is also true for $p+1:$
$$a_1+a_1r+a_1r^2+a_1r^3=a_1\frac{1-r^4}{1-r}$$
etc.
A: The case $a_n=2^n$ is the most intuitive to me because:
$$2^n=2^n(2-1)=2^{n+1}-2^n$$
Then noticing the telescoping sum, the formula easily follows.
If we try to do the same kind off thing with $x$ instead of $2$ we get,
$$x^n=x^n\frac{x-1}{x-1}=\frac{x^{n+1}-x^n}{x-1}$$
$$=\frac{x^{n+1}}{x-1}-\frac{x^n}{x-1}$$
And again the result easily follows by telescoping.
