# 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 add fractions by using a common denominator. $\qquad$ – Michael Hardy May 2 '16 at 20:45
• The proposed proposition is not true if $r=1$. There's no need to assume $r\notin(0,1)$, but you do need to assume $r\ne1$. $\qquad$ – Michael Hardy May 2 '16 at 20:46
• Oh, the initial statement is wrong, $\sum\dots=\frac{1-r^{n+1}}{1-r}$ – Simply Beautiful Art May 2 '16 at 20:47

$$\frac{1-r^{k+1}}{1-r}=\frac{1-r^{k}}{1-r}+r^{k+1}$$
You've got $\displaystyle \frac{1-r^k}{1-r} + r^k$. To add these, you can use $1-r$ as a common denominator. After that, routine simplifications do the rest.
As a side note, you can also do this without induction. I find this method easy to remember, it's short and I usually forget the formula anyway. Set $s_n=1+r+r^2+ \dots +r^n$. Then $rs_n=r+r^2+\dots +r^n+r^{n+1}$. Thus $s_n(1-r)=s_n-rs_n=1-r^{n+1}$. It follows that $s_n=\frac{1-r^{n+1}}{1-r}$.