Sum of Geometric Sequence/Proof 2

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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.

$\blacksquare$


Sources

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