Sum of Sequence of Power by Index/Proof 1
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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
\(\ds \sum_{j \mathop = 0}^n j x^j\) | \(=\) | \(\ds x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} x^j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \sum_{0 \mathop \le j \mathop \le n} j x^j} - n x^{n + 1} + \paren {\sum_{0 \mathop \le j \mathop \le n - 1} x^{j + 1} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - 1} \sum_{0 \mathop \le j \mathop \le n} j x^j\) | \(=\) | \(\ds n x^{n + 1} - \paren {\sum_{1 \mathop \le j \mathop \le n} x^j}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n + 1} - \paren {\sum_{0 \mathop \le j \mathop \le n} x^j} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n x^{n + 1} - \paren {\frac {x^{n - 1} - 1} {x - 1} } + 1\) | Sum of Geometric Sequence | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{j \mathop = 0}^n j x^j\) | \(=\) | \(\ds \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}\) |
$\blacksquare$