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\(\[1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\]\)Compute 

Guest Nov 8, 2018
 #1
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it's so urgent you couldn't post the image so we could see it!

Rom  Nov 8, 2018
 #2
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\(\[1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\]\)

Guest Nov 8, 2018
 #3
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∑[n/(2^n), n, 1, ∞] =2

Here is the partial sum formula:

sum_(n=1 to m) = n/2^n = 2^(-m) (-m + 2^(m + 1) - 2)

Guest Nov 8, 2018
edited by Guest  Nov 8, 2018
 #4
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Here, we can use \(\frac{n}{1-r}\)  , so  \(\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=\boxed{2}\) .

tertre  Nov 9, 2018

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