Does formula exist for this sum?

by db453r
Tags: exist, formula
db453r is offline
Sep12-13, 11:49 AM
P: 3

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.
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jedishrfu is offline
Sep12-13, 12:18 PM
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there is a PF discussion of this series that may help:
D H is online now
Sep12-13, 12:23 PM
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Quote Quote by db453r View Post
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, ?

db453r is offline
Sep12-13, 12:32 PM
P: 3

Does formula exist for this sum?

Oops. Yeah, that's what I meant.
db453r is offline
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 is offline
Sep15-13, 08:51 AM
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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 (**):
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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