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This question already has an answer here:

Evaluate $$\sum\limits_{k=1}^{n} \frac{k}{2^k}$$

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marked as duplicate by user147263, Thomas Andrews, dustin, apnorton, Daniel W. Farlow Feb 19 '15 at 3:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I've posted answers to essentially this same question several times. This question seems to get asked more frequently than all others. $\endgroup$ – Michael Hardy Feb 19 '15 at 2:30
  • $\begingroup$ I'm sorry. I'm Polish. How to find? $\endgroup$ – sasza90 Feb 19 '15 at 2:32
  • $\begingroup$ This is his first, he does not know that it is a duplicate. Anyway he is not asking anything, just posting a problem. $\endgroup$ – YTS Feb 19 '15 at 2:32
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    $\begingroup$ This is "$\infty$", is to be "$n$" $\endgroup$ – sasza90 Feb 19 '15 at 2:39
  • $\begingroup$ Although the question to which it was an answer isn’t actually a duplicate, this answer completely answers the present question using only very elementary techniques. This answer was for the infinite series, but the same technique works here. $\endgroup$ – Brian M. Scott Feb 19 '15 at 2:40
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HINT: Let $$S = \sum_{k = 1}^{n}(\frac{k}{2^k})$$

Try evaluating $S - \frac{S}{2}$ and you will see the trend.

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You have that:

$$\displaystyle\sum_{k=0}^n r^k=\displaystyle\frac{1-r^{n+1}}{1-r}$$

derive the both side to get an expression to this sum, observe that your first term is $1$ not zero. Take $r=\displaystyle\frac{1}{2}$.

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$$\sum\limits_{k=1}^{n} \frac{k}{2^k} = \frac{1}{2} \sum_{k=1}^{n} \frac{k}{2^{k-1}} = \frac{1}{2} \left(\sum_{k=1}^{n} \frac{k - 1}{2^{k-1}} + \sum_{k=1}^{n} \frac{1}{2^{k-1}}\right) = ....$$

Good?

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  • $\begingroup$ Good or wrong? I'm not sure. $\endgroup$ – sasza90 Feb 19 '15 at 13:03

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