Induction to prove that for any $r \in \mathbb{R}$ such tht $r \notin (0,1)$ $\sum_{i=1}^n r^i-1 = \frac{(1-r^n)}{1-r}$ for all $n \in \mathbb{N}$. Use induction to prove that for any $r \in \mathbb{R}$ such that $r \notin (0,1)$ $$\sum_{i=1}^n r^{i-1} = \frac{1-r^n}{1-r}$$ for all $n \in \mathbb{N}$.
My method:
Assume $$\sum_{i=1}^k r^{i-1} = \frac{1-r^k}{1-r}$$
Then show $$\sum_{i=1}^{k+1} r^{i-1} = \frac{1-r^{k+1}}{1-r}$$ 
Then $$\sum_{i=1}^{k+1} r^{i-1} = \sum_{i=1}^k r^{i-1} + r^{(k+1)-1}$$
$$= \frac{1-r^k}{1-r} + r^k$$
I think I am very close I'm not sure how to finish it though.

You want to show
$$\frac{1-r^{k+1}}{1-r}=\frac{1-r^{k}}{1-r}+r^{k+1}$$
Simple algebra should get you there.
