Prove by induction $1 + z + z^2 + .... +z^n = \frac{z^{n+1}-1}{z-1}$ for $z \neq$ 1 Prove $$1 + z + z^2 \ldots +z^n = \frac{z^{n+1}-1}{z-1} \quad \text{ where } z\ne 1 $$
So I started to prove this by induction.
What have I done so far:
Let $n= 1$
$$1+ z + z^2 \ldots+z^n = \frac{z^{n+1}-1}{z-1} \\
1+z \stackrel {?}{=}\frac{z^{2}-1}{z-1} \\~\\
1+z \stackrel {?}{=} \frac{(z+1)(z-1)}{z-1} \\
1+ z \stackrel{?}{=} z+1 \quad \checkmark$$
But I think this is wrong? And I don't know what to do next
and how to do it? But for (a) and (b) use
$e^{i\theta}$ and go from there. Or is this proof an assumption?
Actual question:

Prove that if $z \ne 1$, then
$$
1+z +z^{2}+\cdots+z^{n}=\frac{z^{n+1}-1}{z-1}
$$
Use this result and De Moivre's formula to establish the
following identities.
(a) $1+ \cos \theta+\cos 2 \theta+\cdots+\cos n\theta=\dfrac{1}{2}+\dfrac{\sin\left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)}$
(b) $\sin \theta+ \sin 2 \theta+\cdots+\sin n
\theta=\dfrac{\sin (n \theta / 2) \sin ((n+1) \theta / 2)}{sin (\theta / 2)}$,
where $0<\theta<2 \pi$.

 A: This is provable using standard algebra; however, if you wish to do this by induction:
For ${n=1}$, we get ${1 + z = \frac{z^2-1}{z-1}=z+1}$, so it works. Now assume
$${1 + z + z^2 + ... + z^k = \frac{z^{k+1}-1}{z-1}}$$
This would imply that
$${1 + z + z^2 + ... + z^{k} + z^{k+1} = \frac{z^{k+1}-1}{z-1} + z^{k+1}}$$
Now we simplify the right hand side
$${=\frac{z^{k+1}-1}{z-1} + \frac{z^{k+1}(z-1)}{z-1} = \frac{z^{k+2}-1}{z-1}}$$
Altogether that's
$${1 + z + z^2 + ... + z^k + z^{k+1} = \frac{z^{k+2}-1}{z-1}}$$
which is the correct formula. And so we have shown that if it works for ${n=k}$, that would imply it works for ${n=k+1}$. We have shown it works for ${n=1}$, and so it works for ${n=2}$ then ${n=3}$... so by the Principle of Mathematical Induction,
$${1 + z + z^2 + z^3 + ... + z^{n} = \frac{z^{n+1}-1}{z-1}}$$
as required.

Algebra proof
Thought I'd add the Algebra proof because why not.
Define
$${S_n = 1 + z + z^2 + ... + z^n}$$
Then you see this is
$${= 1 + z(1 + z + z^2 + ... + z^{n-1})}$$
But
$${1 + z + z^2 + ... + z^{n-1} = (1 + z + z^2 + ... + z^n)-z^n = S_n-z^n}$$
So we get overall
$${S_n = 1 + z(S_n - z^n)}$$
We can now rearrange for ${S_n}$;
$${\Rightarrow S_n(1-z) = 1-z^{n+1}}$$
And so
$${S_n = \frac{1-z^{n+1}}{1-z}}$$
This looks different from your form, but it's the same since we just multiplied by ${\frac{-1}{-1}}$:
$${\Rightarrow S_n = \frac{1-z^{n+1}}{1-z}\times \frac{-1}{-1} = \frac{z^{n+1}-1}{z-1}}$$
And so
$${1 + z + z^2 + ... + z^n = \frac{z^{n+1}-1}{z-1}}$$

Part (a) and (b)
Indeed for part (a) and (b) you want to use the fact that
$${(e^{i\theta})^{n} = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)}$$
After applying this formula with ${z=e^{i\theta}}$, you want to equate both real and imaginary parts on the left and right hand sides of the formula. Can you take it from here?
A: So you have done the “base case” right, but it is easier to start from $n=0$ and then you have $1 = (z - 1)/(z-1)$ which is obviously correct for all $z \ne 1$.
For the “inductive step” you want to start from the assumption that this is true for some $n$ and prove the corresponding function for $n+1$ from that assumption. So you have that
$$
1 + \dots + z^n = \frac{z^{n+1}-1}{z-1}
$$
as a given assumption and you want to prove that if you add $z^{n+1}$ to both sides that you get $(z^{n+2} - 1)/(z-1)$ on the right-hand side.
This is very easy to do by writing $z^{n+1}$ as $(z^{n+2} - z^{n+1})/(z-1)$ and then adding the numerators, again making use of the assumption that $z \ne 1$.
Indeed the second part is I think to substitute in $e^{i\theta}$ and then take either the real or imaginary part of the resulting expression to get the correct sums on the left-hand side.
A: You should start from what you know, not from what you want to prove.  For $n=1$ you can say $1+z=1+z\cdot \frac {z-1}{z-1}=\frac{z^2-1}{z-1}$.  For the inductive step, see Riemann'sPointyNose's answer
A: In formula
$$a^{n+1}-b^{n+1}=(a-b)(a^{n} + ba^{n-1}+ ... + b^{n})$$
put $z=a, b=1$. For formula itself you can use induction, or simply open brackets.
