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Does formula exist for this sum?

by db453r
Tags: exist, formula
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db453r
#1
Sep12-13, 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.
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jedishrfu
#2
Sep12-13, 12:18 PM
P: 2,807
there is a PF discussion of this series that may help:

http://www.physicsforums.com/showthread.php?t=220640
D H
#3
Sep12-13, 12:23 PM
Mentor
P: 15,066
Quote Quote by db453r View Post
[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 ?

db453r
#4
Sep12-13, 12:32 PM
P: 3
Does formula exist for this sum?

Oops. Yeah, that's what I meant.
db453r
#5
Sep12-13, 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]
arildno
#6
Sep15-13, 08:51 AM
Sci Advisor
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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^{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


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