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Math Help - arithmetic-geometric series

  1. #1
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    arithmetic-geometric series

    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?
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  2. #2
    Junior Member AlvinCY's Avatar
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    I think you'll find this useful:

    PlanetMath: arithmetic-geometric series
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  3. #3
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    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}

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