# Sum of Geometric Sequence/Proof 2

## Theorem

Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.

Let $n \in \N_{>0}$.

Then:

$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$

## Proof

Let $\displaystyle S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.

Then:

 $\displaystyle \paren {x - 1} S_n$ $=$ $\displaystyle x S_n - S_n$ $\displaystyle$ $=$ $\displaystyle x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j$ $\displaystyle$ $=$ $\displaystyle x^n + \sum_{j \mathop = 1}^{n - 1} x^j - \paren {x^0 + \sum_{j \mathop = 1}^{n - 1} x^j}$ $\displaystyle$ $=$ $\displaystyle x^n - x^0$ $\displaystyle$ $=$ $\displaystyle x^n - 1$

The result follows.

$\blacksquare$