
#1
Sep1213, 11:49 AM

P: 3

[itex]\sum_{i=1}^{n}[i/2^i][/itex]
Have looked and looked and cannot find it anywhere. EDITED: To correct mistake. 



#2
Sep1213, 12:18 PM

P: 2,490

there is a PF discussion of this series that may help:
http://www.physicsforums.com/showthread.php?t=220640 



#3
Sep1213, 12:23 PM

Mentor
P: 14,471

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



#4
Sep1213, 12:32 PM

P: 3

Does formula exist for this sum?
Oops. Yeah, that's what I meant.




#5
Sep1213, 12:46 PM

P: 3

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
Sep1513, 08:51 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

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^{i1}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 


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