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Find the sum of the series

  1. May 16, 2005 #1
    I need to find the sum of:

    [tex]\sum_{n=2}^\infty\frac{50(-2)^{n-1}3^{n+2}}{7^n}[/tex]


    the only series that we've been taught to add up is geometric. the above series is not geometric,is it?
     
  2. jcsd
  3. May 16, 2005 #2

    OlderDan

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    Whatever it is, it has been disquised. Try gouping all the terms you can together to the power n, and reduce everything else and see what you get.
     
  4. May 16, 2005 #3
    is [tex]\frac{-8100}{91}[/tex] correct?
     
  5. May 16, 2005 #4

    OlderDan

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    Looks OK to me
     
  6. May 17, 2005 #5
    how would you compute this series by hand? i did this on a calculator. I don't see how i can combine the exponents given that they have different bases
     
  7. May 17, 2005 #6

    OlderDan

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    [tex]\sum_{n=2}^\infty\frac{50(-2)^{n-1}3^{n+2}}{7^n}[/tex]

    [tex]\frac{50 \bullet 9}{-2}\sum_{n=2}^\infty\frac{(-2)^{n}3^{n}}{7^n}[/tex]

    [tex]\frac{50 \bullet 9}{-2}\sum_{n=2}^\infty\left(\frac{-6}{7}\right)^n[/tex]

    [tex]\frac{50 \bullet 9}{-2}\left(\frac{-6}{7}\right)^2\sum_{n=0}^\infty\left(\frac{-6}{7}\right)^n[/tex]

    [tex]\frac{-50 \bullet 27}{49}\sum_{n=0}^\infty\left(\frac{-6}{7}\right)^n[/tex]

    If you have learned to do geometric sums, you can finish it. If you don't see how I changed the sum from n = 2 to n = 0, write out the first few terms of the sum to see how you can factor out the squared term.
     
  8. May 17, 2005 #7
    beautiful. thank you!
     
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