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Thread: alternating geometric sequence manipulation

  1. #1
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    Question alternating geometric sequence manipulation

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

    help please am I allowed to multiply by $\displaystyle \frac{-1}{4}^{-2}$
    Last edited by Intsecxtanx; Nov 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: .$\displaystyle \sum _{n=1}^{\infty} \frac{2}{3}\left(\text{-}\frac{1}{4}\right)^{n+1} $

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

    We have: .$\displaystyle 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, $\displaystyle a \,=\,\frac{2}{3}\left(\text{-}\frac{1}{4}\right)^2\,=\,\frac{1}{24}$
    . . and common ratio, $\displaystyle r \,=\,\text{-}\frac{1}{4}$

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

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