Prove that $1+a+a^2+...+a^n$ = $\frac{1-a^{n+1}}{1-a}$ for all $a\ne 1$.

Question

I've started by saying $a_n$=$1+a+a^2+...+a^n$=$\frac{1-a^{n+1}}{1-a}$

Now I think I need to do: $lim_{n\to \infty}$$\frac{1-a^{n+1}}{1-a}$

Should let $a=2$ and then solve the limit, then write a proof for it using $\epsilon$?

I've also considered that $1+a+a^2+...+a^n$=$1+a^n$. But I'm not sure what to do with this information.

I'd like to understand how to start this problem properly.

Answer 1

Hint: You can multiply out $$(1+a+a^2+…+a^n)(1-a)=…$$

Answer 2

Or, a standard proof by induction.

$1+a+a^2+...+a^n =\dfrac{1-a^{n+1}}{1-a} $ is true for $n=0$ and $n=1$.

If true for $n$, then

$\begin{array}\\ 1+a+a^2+...+a^n+a^{n+1} &=(1+a+a^2+...+a^n)+a^{n+1}\\ &=\dfrac{1-a^{n+1}}{1-a}+a^{n+1}\\ &=\dfrac{1-a^{n+1}+a^{n+1}(1-a)}{1-a}\\ &=\dfrac{1-a^{n+1}+a^{n+1}-a^{n+2}}{1-a}\\ &=\dfrac{1-a^{n+2}}{1-a}\\ \end{array} $

which completes the proof.