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How this summation is to 3?

  1. Oct 8, 2009 #1
    Hi everyone.
    I hardly remember the fomulas of summation of sequence.

    I got this problem.

    [tex]{\frac{1}{8}}\sum^{\infty}_{n=2}n({\frac{3}{4}})^{n-2}[/tex]

    The result is 2.5.
    How can I solve this problem?

    Thanks all. :)
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 8, 2009 #2
    Assuming |r| < 1 then

    [tex]\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}[/tex]

    Differentiation both sides with respect to r gives:

    [tex]\sum_{n=1}^{\infty} n \cdot r^{n-1} = \frac{1}{(1-r)^2}[/tex]

    This should give you a push in the right direction.

    (Warning: Be careful of your initial index.)

    --Elucidus
     
  4. Oct 9, 2009 #3
    You suggested me very good approach.
    However, the problem still remains,,,

    my equation is n vs (n-2), not n vs (n-1)

    Thanks!
     
  5. Oct 9, 2009 #4

    Office_Shredder

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    Raising it to the power of n-2 instead of n-1 is just dividing it by 3/4. You should be able to find a way to modify your series so that you have an n-1 in the exponent
     
  6. Oct 9, 2009 #5
    You are absolutely right.
    I was so stupid.

    Thank you ;-)
     
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