Does formula exist for this sum?

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[itex]\sum_{i=1}^{n}[i/2^i][/itex]

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.
 
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  • #3
D H
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[itex]\sum_{i=1}^{n}[n/2^n][/itex]
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 looked and looked and cannot find it anywhere.
Have you tried Wolfram alpha, www.wolframalpha.com ?
 
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  • #4
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Oops. Yeah, that's what I meant.
 
  • #5
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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]
 
  • #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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