Induction help with final answer

Question

Use induction to prove that for any complex number $z$ that does not equal $1$ and integer n is greater or equal to 1: $$ 1+z+z^2+...+z^n = \frac{1-z^{n+1}}{1-z} $$

So far for the base case I used $z=2$ and $n=1$ but I can't seem to get it to match the left hand side.

For the rest of the solution I have let $k=n$ $$ 1+z+z^2+...+z^k=\frac{z^{k+1}-1}{z-1} $$ Now for $k+1$: $$ 1+\cdots +z^k+z^{k+1}=\frac{z^{k+1}-1}{z-1}\\ \frac{z^{k+1}-1}{1-z} + z^{1+k}\\ \frac{z^{k+1}-1+z^{k+1}-z^{k+1}}{z-1} \text{(found a common denominator)}\\ -1+z^{k+2}/{z-1}. $$ This is my final spot that I'm stuck at.

I'm not sure if I added the exponents correctly the $z^{k+2}$ is not in the original numerator therefore my left side doesn't match my right Side so I don't know where I went wrong.

Answer 1

The base case must apply to all $z$. Take $n =1$ and work out from there.

$$1+z^1 = \dfrac{1-z^2}{1-z} = \dfrac{(1-z)(1+z)}{(1-z)}$$

(The base case can actually be taken to be $n = 0$, which holds: $z^0 = 1 = \dfrac{1-z^1}{1-z}$)

Note that multiplying your numerator and denominator by $(-1)$ gives $$\dfrac{-1 + z^{k+2}}{z-1} = \dfrac{1-z^{(k+1)+1}}{1-z}$$ as desired. I.e., you arrived at what you need to show.

Answer 2

For $n=k+1$, $$ 1+z+z^2+...+z^k+z^{k+1}=\frac{1-z^{k+1}}{1-z}+ z^{k+1} =\dfrac{1- z^{k+1} +z^{k+1}-z^{k+2}}{1-z}= \frac{1-z^{k+2}}{1-z}. $$