# Sum of Arithmetic-Geometric Sequence

## 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 1

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

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

### Basis for the Induction

$\map P 1$ is the case:

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

demonstrating that $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\ds \sum_{k \mathop = 0}^m \paren {a + k d} r^k$ $=$ $\ds \sum_{k \mathop = 0}^{m - 1} \paren {a + k d} r^k + \paren {a + m d} r^m$ $\ds$ $=$ $\ds \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2} + \paren {a + m d} r^m$ Induction Hypothesis $\ds$ $=$ $\ds \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2}$ $\ds$  $\, \ds + \,$ $\ds \frac {a r^m \paren {1 - r} } {1 - r} + \frac {m d r^m \paren {1 - r}^2} {\paren {1 - r}^2}$ common denominator (2 instances) $\ds$ $=$ $\ds \frac {a \paren {1 - r^m + r^m \paren {1 - r} } } {1 - r}$ $\ds$  $\, \ds + \,$ $\ds \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} + r d \paren {m r^{m - 1} \paren {1 - r}^2} } {\paren {1 - r}^2}$ simplifying $\ds$ $=$ $\ds \frac {a \paren {1 - r^m + r^m - r^{m + 1} } } {1 - r}$ $\ds$  $\, \ds + \,$ $\ds \frac {r d \paren {1 - m r^{m - 1} + m r^m - r^m + m r^{m - 1} - 2 m r^m + m r^{m + 1} } } {\paren {1 - r}^2}$ multiplying out $\ds$ $=$ $\ds \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - m r^m - r^m + m r^{m + 1} } } {\paren {1 - r}^2}$ cancelling out terms $\ds$ $=$ $\ds \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {m + 1} r^m + m r^{m + 1} } } {\paren {1 - r}^2}$ simplification

So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

$\blacksquare$

## Proof 2

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

$\blacksquare$

## Linguistic Note

In the context of an arithmetic sequence or arithmetic-geometric sequence, the word arithmetic is pronounced with the stress on the first and third syllables: a-rith-me-tic, rather than on the second syllable: a-rith-me-tic.

This is because the word is being used in its adjectival form.