Sum of Arithmetic-Geometric Sequence

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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.

$\blacksquare$


Also see


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.


Sources