# Does formula exist for this sum?

by db453r
Tags: exist, formula
 P: 3 $\sum_{i=1}^{n}[i/2^i]$ Have looked and looked and cannot find it anywhere. EDITED: To correct mistake.
 P: 2,493 there is a PF discussion of this series that may help: http://www.physicsforums.com/showthread.php?t=220640
Mentor
P: 14,476
 Quote by db453r $\sum_{i=1}^{n}[n/2^n]$
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 ?

P: 3

## Does formula exist for this sum?

Oops. Yeah, that's what I meant.
 P: 3 Wow. Didn't know Wolfram could do that. Thanks. Here's what it gave me: $\sum_{i=0}^{n} i/2^{-i} = 2^{-n}(-n+2^{n+1} -2)$
 Sci Advisor HW Helper PF Gold P: 12,016 You can solve this by hand by using a neat trick. Form the auxiliary function (*): $$F(x)=\sum_{i=1}^{i=n}(\frac{x}{2})^{i}$$, that is, F(x) is readily seen to be related to a geometric sum, with alternate expression (**): $$F(x)=\frac{1-(\frac{x}{2})^{n+1}}{1-\frac{x}{2}}-1$$ Now, the neat trick consists of differentiating (*), and we get: $$F'(x)=\sum_{i=1}^{i=n}i*x^{i-1}2^{-i}$$ that is, we have: $$F'(1)=\sum_{i=1}^{i=n}i*2^{-i}$$ 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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