Sum of Geometric Sequence/Proof 4

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

Lemma

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

Then:

$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$

$\Box$

Then by the lemma:

\(\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i\)

\(=\)

\(\ds 1 - x^n\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}\)

\(=\)

\(\ds \frac {x^n - 1} {x - 1}\)

$\blacksquare$