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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lfryerda | High School Teacher | (Level 2) Educator

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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.

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