Proof a Sum with real numbers

Question

I have to prove with Induction that:

$ \sum_{i=1}^n a^{i-1} = \frac{a^n -1 }{a-1}$ where $a \in R $ \ {0,1} with $a^0$ = 1

In the first induction step I get to divide by 0, because for a=1 it's 1-1. Well I think I didn't quite understand the parameters, I guess.

$a$ can be any real number; $n$ has to be a positive integer. You can only do induction over the integers, not real numbers. (Well, that's not 100% true — there is a version of "induction" that works over the real numbers, but it's a bit more complicated and much less well-known.) — Akiva Weinberger, Oct 11 '15 at 15:38

score 1 · Accepted Answer · answered Oct 11 '15 at 15:53

Looks like someone in the comments pointed out the confusion between $a$ and $n$. With that cleared up, you need to prove the base case; show that the equality holds for $n=1$. Once that's done, make your induction hypothesis:

Suppose $\sum_{i=0}^{n-1} a^i = \frac{a^n-1}{a-1}$ holds for all $n \in \{1,2,\ldots , k\}$.

Now you want to show that equality holds for $n = k+1$, so evaluate $$\sum_{i=0}^{(k+1)-1} a^i = \sum_{i=0}^{k} a^i $$ The trick here would be to write this quantity in a meaningful way (one that lets you apply the induction hypothesis). So in this case I'd recommend writing $$\sum_{i=0}^{k} a^i = \left(\sum_{i=0}^{k-1} a^i \right)+ a^k $$ You are in a position to say something about $\sum_{i=0}^{k-1} a^i$ with the induction hypothesis. Can you proceed from here?

Ok so now iam at a point where I have to prove that $ \frac{2a^n + a -2}{a-1} = \frac{a^{n+1}-1}{a-1}$ but I don't know how to get there... — Maximilian Kindshofer, Oct 11 '15 at 15:56
I think you have made an algebra error. You should find $$\frac{a^k-1}{a-1}+a^k = \frac{a^k-1}{a-1}+a^k\frac{a-1}{a-1} $$ — graydad, Oct 11 '15 at 15:59

Proof a Sum with real numbers

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