Sum of Arithmetic-Geometric Sequence/Proof 2

Theorem

Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:

$a_k = \paren {a + k d} r^k$ for $k = 0, 1, 2, \ldots, n - 1$

Then its closed-form expression is:

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

Proof

\(\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k\)

\(=\)

\(\ds a \sum_{k \mathop = 0}^{n - 1} r^k + d \sum_{k \mathop = 0}^{n - 1} k r^k\)

\(\ds \)

\(=\)

\(\ds \frac {a \paren {1 - r^n} } {1 - r} + d \sum_{k \mathop = 0}^{n - 1} k r^k\)

Sum of Geometric Sequence

\(\ds \)

\(=\)

\(\ds \frac {a \paren {1 - r^n} } {1 - r} + d \paren {\frac {\paren {n - 1} r^{n + 1} - n r^n + r} {\paren {r - 1}^2} }\)

Sum of Sequence of Power by Index

Hence the result, after algebra.

$\blacksquare$