# Evaluate $\sum\limits_{k=1}^{n} \frac{k}{2^k}$ [duplicate]

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

• I've posted answers to essentially this same question several times. This question seems to get asked more frequently than all others. – Michael Hardy Feb 19 '15 at 2:30
• I'm sorry. I'm Polish. How to find? – sasza90 Feb 19 '15 at 2:32
• This is his first, he does not know that it is a duplicate. Anyway he is not asking anything, just posting a problem. – YTS Feb 19 '15 at 2:32
• This is "$\infty$", is to be "$n$" – sasza90 Feb 19 '15 at 2:39
• 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. – Brian M. Scott Feb 19 '15 at 2:40

HINT: Let $$S = \sum_{k = 1}^{n}(\frac{k}{2^k})$$

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

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 $r=\displaystyle\frac{1}{2}$.

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

Good?

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