- #1

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

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.

Last edited:

- Thread starter db453r
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- #1

- 3

- 0

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.

Last edited:

- #2

jedishrfu

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https://www.physicsforums.com/showthread.php?t=220640

- #3

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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##.[itex]\sum_{i=1}^{n}[n/2^n][/itex]

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

Have you tried Wolfram alpha, www.wolframalpha.com ?Have looked and looked and cannot find it anywhere.

- #4

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Oops. Yeah, that's what I meant.

- #5

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Here's what it gave me:

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

- #6

arildno

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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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