# Does formula exist for this sum?

1. ### db453r

3
$\sum_{i=1}^{n}[i/2^i]$

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.

Last edited: Sep 12, 2013

### Staff: Mentor

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 you tried Wolfram alpha, www.wolframalpha.com ?

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4. ### db453r

3
Oops. Yeah, that's what I meant.

5. ### db453r

3
Wow. Didn't know Wolfram could do that. Thanks.

Here's what it gave me:

$\sum_{i=0}^{n} i/2^{-i} = 2^{-n}(-n+2^{n+1} -2)$

6. ### arildno

12,015
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:
$$F'(1)=\sum_{i=1}^{i=n}i*2^{-i}$$