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Finding the Sum of a particular Power Series

  1. Nov 2, 2004 #1
    I need to obtain the sum of the following series

    [tex] \sum _{n=1} ^{\infty} \frac{n^2}{2^n} [/tex]

    Well, with the aid of Mathematica, I get the answer, which is 6. What I'm trying to do now is work my way backwards from there. I need to express it through the geometric series

    [tex] \sum _{n=0} ^{\infty} x^n = \frac{1}{1-x} [/tex]

    In fact, my guess is that I should use

    [tex] \sum _{n=1} ^{\infty} n x^{n-1} = \frac{1}{\left( x - 1 \right) ^2} [/tex]

    but it can't get it to fit in.

    Thank you very much.
  2. jcsd
  3. Nov 2, 2004 #2


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    [tex]S = \sum_{n=0}^{\infty} x^n[/tex]


    [tex]\sum_{n=0}^{\infty}n^2 x^n = x \frac {d}{dx} x \frac {dS}{dx}[/tex]
  4. Nov 2, 2004 #3
    Thank you so much. This definitely works out!!!
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