# Sum of Geometric Sequence/Proof 3

## 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:

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

## Proof

$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$

Set $a = x$ and $b = 1$:

$\ds x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = \paren {x - 1} \sum_{j \mathop = 0}^{n - 1} x^j$

from which the result follows directly.

$\blacksquare$