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:

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

Proof

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

Then:

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

The result follows.

$\blacksquare$