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.
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:
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?