Sum of Geometric Sequence/Corollary 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:
- $\ds \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$
Proof
\(\ds \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}\) | \(=\) | \(\ds \sum_{j \mathop = 0}^{n - 1} D_x x^j\) | Power Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds D_x \sum_{j \mathop = 0}^{n - 1} x^j\) | Sum Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds D_x \frac {x^n - 1} {x - 1}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {n x^{n - 1} } \paren {x - 1} - \paren 1 \paren {x^n - 1} } {\paren {x - 1}^2}\) | Quotient Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{j \mathop = 0}^{n - 1} j x^j\) | \(=\) | \(\ds x \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}\) | Distributive Property | ||||||||||
\(\ds \) | \(=\) | \(\ds x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}\) | by the above result | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}\) |
$\blacksquare$