arithmetic-geometric series

**kcsteven** · February 5th 2007, 05:01 PM

The general arithmetic-geometric series of n terms is a series of the form:
a + (a+d)r + (a + 2d)r^2 + ... +[a + (n - 1)d]r^n-1 Where the first factors of each term form an arithmetic series and the second factors form a geometric series. What technique could I use to derive a formula for the sum of this series? Could I let S denote the sum of the general geometric series of n terms. Multiply S by r and write the series corresponding to the difference between S and rS. How would I use this expression to deducde the formula for the sum of the general geometric series?

**AlvinCY** · February 5th 2007, 07:02 PM

I think you'll find this useful:

PlanetMath: arithmetic-geometric series

**Soroban** · February 5th 2007, 09:52 PM

Hello, Keith!

The general arithmetic-geometric series of $n$ terms is a series of the form:
. . $S \:=\:a + (a+d)r + (a + 2d)r^2 + \cdots +[a + (n - 1)d]r^{n-1}$
where the first factors of each term form an arithmetic series
and the second factors form a geometric series.

What technique could I use to derive a formula for the sum of this series?
Could I let $S$ denote the sum of the general geometric series of $n$ terms?
Multiply $S$ by $r$ and then write the series corresponding
. . to the difference between $S$ and $rS$ ? . Yes ... a great game plan!

How would I use this to deduce the formula for the sum of the general series?

The original series:
. . . $S\:=\:a + (a+d)r + (a+2d)r^2 + (a+3a)r^3 + \cdots + [a+(n-1)d]r^{n-1}$

Multiply by $r\!:$
. . . $rS\:=\qquad\quad ar \quad+ \quad(a+d)r^2 + (a+2d)r^3 + \cdots$ $+ [a+(n-2)d]r^{n-1} + [a + [n-1]d]r^n$

Subtract: . $S - rS\:=\;a + \underbrace{dr + dr^2 + dr^3 + \cdots + dr^{n-1}}_{\text{geometric series}} - [a + (n-1)d]r^n$

. . The geometric series has first term $dr$ , common ratio $r$ , and $n-1$ terms.
. . . . Its sum is: . $dr\,\frac{1-r^{n-1}}{1-r}$

So we have: . $(1 - r)S\;=\;a +dr\,\frac{1 - r^{n-1}}{1-r} - [a + (n-1)d]r^n$

. . $(1-r)S\:=\:\frac{a(1-r) + dr(1 - r^{n-1}) - (1-r)[a + (n-1)d]r^n}{1 - r}$

Therefore: . $S \:=\:\frac{a(1-r) + dr(1 - r^{n-1}) - (1-r)[a + (n-1)d]r^n}{(1-r)^2}$

Thread: arithmetic-geometric series

LinkBack

Thread Tools

Display

arithmetic-geometric series

Similar Math Help Forum Discussions

sum of non-arithmetic,non-geometric series

Need help with arithmetic/geometric series

Arithmetic / Geometric series HELP!

Geometric and Arithmetic series question

The general arithmetic-geometric series

Search Tags