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

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

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.

  • 0
    The answer is given in http://math.stackexchange.com/a/29024/24682012-04-05

2 Answers 2

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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...

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