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Sum series

  1. Oct 9, 2011 #1

    Suy

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    1. The problem statement, all variables and given/known data

    [PLAIN]http://img651.imageshack.us/img651/2219/unledybi.jpg [Broken]
    can someone teach me how to do it ?
    answer is k=1/2,-7/17
    thanks
    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 13, 2011 #2
    [tex]\sum_{n=1}^∞(3-27k)k^{2n}=(3-27k)\sum_{n=1}^∞(k^2)^n=(3-27k)(\sum_{n=0}^∞(k^2)^n-1)=(3-27k)(\frac{1}{1-k^2}-1)[/tex]
    assuming [itex]|k|<1[/itex]
    [tex](3-27k)(\frac{1}{1-k^2}-1)=-7k[/tex]
    Now multiply with [itex](1-k^2) [/itex] and use your favorite CAS program to get
    [tex]-3(-1+9k)k=7(-1+k^2)[/tex]
    Now solve for k
     
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