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Sum of series

  1. Apr 23, 2005 #1
    Determine the sum of the following series

    [tex]\sum_{n=1}^\infty \frac{2^n+6^n}{9^n}[/tex] or can be written as...
    [tex]\sum_{n=1}^\infty \frac{8^n}{9^n}[/tex]

    [tex] A_1 = 8/9, A_2 = 64/81, A_3 = 512/729 [/tex]

    common ration (r)= 8/9
    first term (a)= 8/9

    so plugging everything i know into the geometric series formula:
    [tex] \frac {a}{1-r}[/tex]

    i get.... 8
    but it's wrong, and i dont see why
     
  2. jcsd
  3. Apr 23, 2005 #2

    saltydog

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    [tex]\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}[/tex]

    And you know what:

    [tex]\frac{k^n}{m^n}[/tex]

    is with k and m constants right?
     
    Last edited: Apr 23, 2005
  4. Apr 23, 2005 #3

    Hurkyl

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    Are you sure 2^n + 6^n = 8^n?
     
  5. Apr 23, 2005 #4


    [tex]\frac{2}{9}\sum_{n=1}^\infty \frac{1^n}{1^n}[/tex] +

    [tex]\frac{6}{9}\sum_{n=1}^\infty \frac{1^n}{1^n}[/tex]


    right?
     
  6. Apr 23, 2005 #5

    Hurkyl

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    I strongly suggest you go back and review the section on exponents in your earlier textbooks -- your problem lies with the fact you don't know how to manipulate them, and it would be much easier to refresh your memory without having to worry about calculus stuff at the same time!
     
  7. Apr 23, 2005 #6
    Testing "2" in for x yields the following statements.

    [tex]2^2 + 6^2 = 40[/tex]
    [tex]8^2 = 64[/tex]

    Hurkyl is right. Rework that step and see what else you can come up with.
     
  8. Apr 23, 2005 #7
    i got 4. final answer regis :tongue2:
    (saltydog did 80% of the work)
     
  9. Apr 23, 2005 #8
    the answer is 2.2857, it was actually an easy question. thanks for the help
     
  10. Apr 23, 2005 #9

    Hurkyl

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    Really? I find it strange that it happened to be a terminating decimal.
     
  11. Apr 24, 2005 #10
    how did you get that? i did [tex]\frac{1}{1- 2/9} + \frac{1}{1-2/3}[/tex] which adds up to 4
     
  12. Apr 24, 2005 #11

    Hurkyl

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    Check the starting index of the sum.
     
  13. Apr 24, 2005 #12
    [tex]\sum_{n=1}^\infty \frac{2^n+6^n}{9^n}[/tex]

    So, that's what we're dealing with, right? I'm not too sharp on series, but I don't thing this is hard.

    Follow saltydog's advice, and the series will become two series,

    [tex]\sum_{n=1}^\infty \frac{2^n}{9^n} + \sum_{n=1}^\infty \frac{6^n}{9^n}[/tex]

    [tex]\sum_{n=1}^\infty (\frac{2}{9})^n + \sum_{n=1}^\infty (\frac{6}{9})^n[/tex]

    Use the geometric series formula on each of those.

    It does?
     
    Last edited by a moderator: Apr 24, 2005
  14. Apr 24, 2005 #13
    sorry 30/7, not 4
     
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