# Prove that `1+r+r^2+cdots+r^n={1-r^{n+1})/{1-r}` for all n is an element of N, when r does not equal 1.

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### 1 Answer

The simplest way to prove this is using induction on `n in N` . For this entire proof, assume that `r ne 1` .

We start by showing the statement is true for n=1. Then:

`LS=1+r`

`RS={1-r^2}/{1-r}` factor numerator

`={(1-r)(1+r)}/{1-r}` cancel common factors

`=1+r`

`=LS`

Now assume the statement is true for n=k, which means that:

`1+r+r^2+cdots+r^k={1-r^{k+1}}/{1-r}`

We then need to show that it is also true for `n=k+1` , so the LS has to equal `RS={1-r^{k+2}}/{1-r}`

`LS=1+r+r^2+cdots+r^k+r^{k+1}` replace with assumption

`={1-r^{k+1}}/{1-r}+r^{k+1}` get common denominator

`={1-r^{k+1}+r^{k+1}(1-r)}/{1-r}` simplify numerator

`={1-r^{k+1}+r^{k+1}-r^{k+2}}/{1-r}` remove extra terms in numerator

`={1-r^{k+2}}/{1-r}`

`=RS`

**By induction, the statement has been proven to be true.**