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Math Help - Summation

  1. #1
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    Sum

    Just can't get this right.
    I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
    I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
    Anyone can help me?
    Last edited by Mimmi; September 12th 2011 at 02:53 PM.
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  2. #2
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    Re: Summation

    Quote Originally Posted by Mimmi View Post
    Just can't get this right.
    I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
    I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
    Anyone can help me?
    \displaystyle \begin{align*} -2 + \frac{4}{3} - \frac{8}{9} + \frac{16}{27} \pm \dots &= -\frac{2}{1} + \frac{4}{3} - \frac{8}{9} + \frac{16}{27} \pm \dots \\ &= -\frac{2^1}{3^0} + \frac{2^2}{3^1} - \frac{2^3}{3^2} + \frac{2^4}{3^3} \pm \dots \\ &= \frac{(-1)^1 \cdot 2^1}{3^0} + \frac{(-1)^2 \cdot 2^2}{3^1} + \frac{(-1)^3 \cdot 2^3}{3^2} + \frac{(-1)^4 \cdot 2^4}{3^3} + \dots \\ &= \frac{(-2)^1}{3^0} + \frac{(-2)^2}{3^1} + \frac{(-2)^3}{3^2} + \frac{(-2)^4}{3^3} + \dots \\ &= \frac{(-2)^{0 + 1}}{3^0} + \frac{(-2)^{1 + 1}}{3^1} + \frac{(-2)^{2 + 1}}{3^2} + \frac{(-2)^{3 + 1}}{3^3} + \dots \\ &= \sum_{n = 0}^{\infty}\frac{(-2)^{n + 1}}{3^n} \end{align*}
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  3. #3
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    Re: Summation

    Quote Originally Posted by Mimmi View Post
    Just can't get this right.
    I have -2+4/3-8/9+16/27... and want to write the sum by useing sigma.
    I guess n is infinity, but after that it stopes for me. Only thing i can see is that i can multiply -2*(-2/3) to get 4/3 and so on. I tryed google to find somewhere i could learn how to think to make the formula but could not find much that help me.
    Anyone can help me?
    Try: \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k 2^k }}{{3^{k - 1} }}}
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  4. #4
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    Re: Summation

    I think it looks like nth term multiplied by (n+1)th term.
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  5. #5
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    Re: Summation

    Quote Originally Posted by arangu1508 View Post
    I think it looks like nth term multiplied by (n+1)th term.
    Wrong. You can get each term by multiplying the previous one by \displaystyle -\frac{2}{3}...
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  6. #6
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    Re: Summation

    Thx! I actually understand how to do this now.
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