Sum of Sequence of Power by Index

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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 1

\(\ds \sum_{j \mathop = 0}^n j x^j\) \(=\) \(\ds x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}\)
\(\ds \) \(=\) \(\ds x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j\)
\(\ds \) \(=\) \(\ds \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} x^j}\)
\(\ds \) \(=\) \(\ds \paren {x \sum_{0 \mathop \le j \mathop \le n} j x^j} - n x^{n + 1} + \paren {\sum_{0 \mathop \le j \mathop \le n - 1} x^{j + 1} }\)
\(\ds \leadsto \ \ \) \(\ds \paren {x - 1} \sum_{0 \mathop \le j \mathop \le n} j x^j\) \(=\) \(\ds n x^{n + 1} - \paren {\sum_{1 \mathop \le j \mathop \le n} x^j}\)
\(\ds \) \(=\) \(\ds n x^{n + 1} - \paren {\sum_{0 \mathop \le j \mathop \le n} x^j} + 1\)
\(\ds \) \(=\) \(\ds n x^{n + 1} - \paren {\frac {x^{n - 1} - 1} {x - 1} } + 1\) Sum of Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds \sum_{j \mathop = 0}^n j x^j\) \(=\) \(\ds \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}\)

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources

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