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Does formula exist for this sum?

  1. Sep 12, 2013 #1
    [itex]\sum_{i=1}^{n}[i/2^i][/itex]

    Have looked and looked and cannot find it anywhere.

    EDITED: To correct mistake.
     
    Last edited: Sep 12, 2013
  2. jcsd
  3. Sep 12, 2013 #2

    jedishrfu

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  4. Sep 12, 2013 #3

    D H

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    That one's easy. There's nothing inside the sum that depends on i, so your sum is the same as ##\frac n {2^n}\sum_{i=1}^n 1##.

    Do you mean ##\sum_{i=1}^n \frac i {2^i}## ?

    Have you tried Wolfram alpha, www.wolframalpha.com ?
     
  5. Sep 12, 2013 #4
    Oops. Yeah, that's what I meant.
     
  6. Sep 12, 2013 #5
    Wow. Didn't know Wolfram could do that. Thanks.

    Here's what it gave me:

    [itex]\sum_{i=0}^{n} i/2^{-i} = 2^{-n}(-n+2^{n+1} -2)[/itex]
     
  7. Sep 15, 2013 #6

    arildno

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    You can solve this by hand by using a neat trick.
    Form the auxiliary function (*):
    [tex]F(x)=\sum_{i=1}^{i=n}(\frac{x}{2})^{i}[/tex], that is, F(x) is readily seen to be related to a geometric sum, with alternate expression (**):
    [tex]F(x)=\frac{1-(\frac{x}{2})^{n+1}}{1-\frac{x}{2}}-1[/tex]
    Now, the neat trick consists of differentiating (*), and we get:
    [tex]F'(x)=\sum_{i=1}^{i=n}i*x^{i-1}2^{-i}[/tex]
    that is, we have:
    [tex]F'(1)=\sum_{i=1}^{i=n}i*2^{-i}[/tex]
    which is your original sum!!

    Thus, you may calculate that sum by differentiating (**) instead, and evaluate the expression you get at x=1
    :smile:
     
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