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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.

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  • $\begingroup$ You add fractions by using a common denominator. $\qquad$ $\endgroup$ – Michael Hardy May 2 '16 at 20:45
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    $\begingroup$ 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 $r\ne1$. $\qquad$ $\endgroup$ – Michael Hardy May 2 '16 at 20:46
  • $\begingroup$ Oh, the initial statement is wrong, $\sum\dots=\frac{1-r^{n+1}}{1-r}$ $\endgroup$ – Simply Beautiful Art May 2 '16 at 20:47
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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.

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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.

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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 $s_n=\frac{1-r^{n+1}}{1-r}$.

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