Sum of Sequence of Power by Index/Proof 2

Theorem

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

for $x \ne 1$.

Proof

From Sum of Arithmetic-Geometric Sequence:

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

Hence:

\(\ds \sum_{j \mathop = 0}^n j x^j\)

\(=\)

\(\ds \frac {0 \paren {1 - x^{n + 1} } } {1 - x} + \frac {x \times 1 \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}\)

putting $a = 0, d = 1$

\(\ds \)

\(=\)

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

initial simplification

\(\ds \)

\(=\)

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

further simplification

Hence the result.

$\blacksquare$