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I am doing a review of stuff from earlier in the semester and I can't prove this by induction:

Use induction on $n$ to verify that $1+x+\cdots+z^n= \frac{1-z^{n+1}}{1-z}$ (for $z\not=1)$. Use this to show that if $c$ is an $n$-th root of $1$ and $c\not=1$, then $1+c+\cdots+c^n=0$.

There is also a follow up question based on that one:

Show that if $c$ is any $n$-th root of $1$ and $c\not=1$ then $1+c+c^2+\cdots+c^{n-1}=0$

Note: If memory serves me correctly there is a misprint in one of these questions. I can't remember which one.

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    @GerryMyerson: I started writing my answer before your 2nd comment. I've deleted it now.2012-03-20

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Induction: $n=1$ is true $1+z={(1-z)(1+z) \over (1-z)}={1-z^2 \over 1-z}$, induction step $n\Rightarrow n+1$ $ 1+z+\dots+z^n+z^{n+1}={1-z^{n+1}\over 1-z} + z^{n+1} ={1-z^{n+1}\over 1-z} + {(1-z)z^{n+1}\over(1-z)} = {1-z^{n+1}+z^{n+1}-z^{n+2}\over 1-z} $

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For the induction, it is true for $n=0:\ 1=\frac{1-z}{1-z}$. Now assume $\sum_{i=0}^k z^i=\frac {1-z^{k+1}}{1-z}$. Then $\sum_{i=0}^{k+1} z^i=\frac {1-z^{k+1}}{1-z}+z^{k+1}=\frac {1-z^{k+1}+z^{k+1}-z^{k+2}}{1-z}=\frac {1-z^{k+2}}{1-z}$