Prove Algebric Identity

Question

Possible Duplicate:
Value of $\sum\limits_n x^n$

Given $a\in\mathbb{R}$ and $0<a<1$

let $(X_n)$ be a sequence defined: $X_n=1+a+a^2+...+a^n$, $\forall n\in\mathbb{N}$.

How do I show that $X_n=\frac{1-a^{n+1}}{1-a}$

Thanks.

Answer 1

Note $aX_n = a + a^2 + \dots + a^n + a^{n+1}$, so that $$X_n - aX_n = 1 - a^{n+1},$$ and then you can probably figure it out from there...

Answer 2

$$\begin{align*} (1-a)(1+a+a^2+\dots+a^n)&=(1+a+\dots+a^n)-a(1+a++\dots+a^n)\\ &=(1+\color{red}{a+\dots+a^n})-(\color{red}{a+a^2++\dots+a^n}+a^{n+1})\\ &=1-a^{n+1} \end{align*}$$