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
\(\ds \) \(=\) \(\ds \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}\)
\(\ds \leadsto \ \ \) \(\ds \sum_{j \mathop = 0}^{n - 1} j x^j\) \(=\) \(\ds x \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}\) Multiplication of Numbers Distributes over Addition
\(\ds \) \(=\) \(\ds x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}\) by the above result
\(\ds \) \(=\) \(\ds \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}\)

$\blacksquare$