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.
$\blacksquare$
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 19$: Arithmetic-Geometric Series: $19.6$