Sum of Geometric Sequence/Corollary 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:

$\ds \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds D_x \frac {x^n - 1} {x - 1}$

$\ds $

$=$

$\ds \frac {\paren {n x^{n - 1} } \paren {x - 1} - \paren 1 \paren {x^n - 1} } {\paren {x - 1}^2}$

$\ds $

$=$

$\ds \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}$

$\ds \leadsto \ \ $

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

$=$

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

$\ds $

$=$

$\ds x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}$

by the above result

$\ds $

$=$

$\ds \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$

$\blacksquare$