Confirm that the identity $1+z+...+z^n=(1-z^{n+1})/(1-z)$ holds for every non-negaive integer $n$ and every complex number $z$, save for $z=1$ Confirm that the identity $1+z+...+z^n=(1-z^{n+1})/(1-z)$ holds for every non-negaive integer $n$ and every complex number $z$, save for $z=1$
I have tried to prove this by induction but I am not sure that I am doing things right, for $ n = 1 $ we have $ (1-z ^ 2) / (1-z) = (1-z) (1+ z) / (1-z) = 1 + z $, then this holds for $ n = 1 $. Suppose now that it holds for $ n $ and see that it holds for $ n + 1 $,$1+z+...+z^n+z^{n+1}=(1-z^{n+1})/(1-z)+z^{n+1}=[(1-z^{n+1})+(1-z)(z^{n+1})]/(1-z)=(1-z^{n+2})/(1-z) $  then this is true for every non-negative integer $ n$. This is OK?
 A: My proof goes like this:
Let $S:=1+z+z^2+...+z^n$
Then
$$zS=z+z^2+z^3+...+z^{n+1}$$
$$zS=S-1+z^{n+1}$$
$$(1-z)S=1-z^{n+1}$$
As $z\neq 1$, it gives the desired result.
A: By induction,
Assume that
$$1+z+z^2+...+z^n=\frac {1-z^{n+1}}{1-z} $$
then
$$1+z+...+z^n+z^{n+1}=$$
$$\frac {1-z^{n+1}}{1-z}+z^{n+1}= $$
$$\frac {1-z^{n+1}+z^{n+1}-z^{n+2}}{1-z} $$
Done.
A: You can also prove that 
$(1-z)\sum_{i=0}^n z^i=$
$\sum_{i=0}^n (1-z)z^i=$ (technically this assummes distribution is known for all finite sums;  that can be proven by induction for the precise and anal).
$\sum_{i=0}^n (z^i-z^{i+1})=$
$\sum_{i=0}^nz^i - \sum_{i=0}^nz^{i+1}= $ (technically we must prove associativity of addition will hold for all finite sums... which is a proof by induction)
$\sum_{i=0}^nz^i - \sum_{j=1}^{n+1}z^{j}= $ (relabel $j = i+1$)
$\sum_{k=0}^nz^i - \sum_{k=1}^{n+1}z^{k}= $ (relabel $k = i$ on the left and $k = j$ on the right)
$(1 + \sum_{k=1}^n z^k) - (\sum_{k=1}^nz^k - z^{n+1}) = $
$1 + (\sum_{i=1}^n z^i - \sum_{i=1}^n z^i) - z^{n+1} =$ (ditto the assumption about associativity.)
$1 - z^{n+1}$.
