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Math Help - alternating geometric sequence manipulation

  1. #1
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    Question alternating geometric sequence manipulation

    find the sum of  \Sigma_{n=1}^{\infty} \frac{2}{3} (  {\frac{-1}{4}}^{n+1} )

    help please am I allowed to multiply by \frac{-1}{4}^{-2}
    Last edited by Intsecxtanx; November 21st 2009 at 03:30 PM.
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  2. #2
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    Hello, Intsecxtanx!

    I hope I've guessed the problem correctly . . .


    Find the sum of: .  \sum _{n=1}^{\infty} \frac{2}{3}\left(\text{-}\frac{1}{4}\right)^{n+1}

    am I allowed to multiply by \left(\text{-}\frac{1}{4}\right)^{-2}
    Well, sort of . . . but why would you do that?

    We have: . S \;=\;\frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^2 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^3 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^4 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^5 + \hdots

    We have a geometric series with first term, a \,=\,\frac{2}{3}\left(\text{-}\frac{1}{4}\right)^2\,=\,\frac{1}{24}
    . . and common ratio, r \,=\,\text{-}\frac{1}{4}

    Substitute into the formula: . S \;=\;\frac{a}{1-r}

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  3. #3
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    Never mind thank you, I get it now, that's a_1 when you plug in n=1 and that's the numerator of the formula for the sum. Thanks!!!!
    Last edited by Intsecxtanx; November 21st 2009 at 07:28 PM.
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