# How is the formula for a finite geometric series found?

I have these two finite geometric series:

$S_n$ = $\sum_{k=0}^n ar^k$

r$S_n$ = $\sum_{k=0}^n ar^{k+1}$

And then we substract both series so:

$S_n$ - r$S_n$ = a - $ar^{n+1}$ //this I understand

$\frac {S_n(1-r)}{1-r}$ = $\frac{a - ar^{n+1}}{1-r}$ //where does 1-r come from?

• $S_n-rS_n=(1-r)S_n$ Dec 8 '15 at 20:52
• @Guest could you elaborate, I don't get why they're equivalent Dec 8 '15 at 20:55
• It's just the distributive property of $\cdot$ with respect to $+$ in $\mathbb{R}$ (more precisely, in this case, right distributivity of multiplication wrt addition). Dec 8 '15 at 20:56

Once you have $$S_n - rS_n = a - ar^{n+1}$$ you can factor both sides: $$S_n ( 1 - r) = a(1 - r^{n+1})$$ (If that factorization isn't immediately obvious to you, just redistribute back through and confirm that you get the original expression again).
Now you can divide both sides by $(1-r)$ and get the desired conclusion.