Does formula exist for this sum?

  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. jedishrfu

    Staff: Mentor

  4. D H

    Staff: Mentor

    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, ?
    1 person likes this.
  5. Oops. Yeah, that's what I meant.
  6. 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. arildno

    arildno 11,265
    Science Advisor
    Homework Helper
    Gold Member

    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 (**):
    Now, the neat trick consists of differentiating (*), and we get:
    that is, we have:
    which is your original sum!!

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