1

Compute the sum

$$\sum_{k=1}^{10}{\dfrac{k}{2^k}}$$

This question is taken from SMO junior (I can't remember which year it is). I have no idea how to start. Can anyone give some hint?

By writing out the sum, one has

$$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ... + \frac{10}{1024}$$

but I don't know how to proceed from here.

Martin Sleziak
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Idonknow
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  • The sum is finite. Just start writing out the terms, find the common denominator among the terms (which is $2^{10}$), and add 'em up. – Mike Pierce Feb 16 '16 at 03:14
  • @idonknow how do you normally add fractions? – Q the Platypus Feb 16 '16 at 03:20
  • if you will have more terms, you can rewrite it as the sum of geometric series. – Kerr Feb 16 '16 at 03:21
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    This is definitely a duplicate... although the posts I can find are for the infinite sum (see e.g. this one, the techniques (differentiation, for instance) are the same, and the topic has been treated numerous times at length. This one is even a direct match. – Clement C. Feb 16 '16 at 03:43

4 Answers4

3

$$\sum_{k=1}^{10}{\dfrac{k}{2^k}}=\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ... + \frac{10}{1024}=\frac{2^9+2*2^8+3*2^7+...+10*2^0}{2^{10}}$$ and notice that the numerator can be written as: $$ \begin{array}{l} 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+ \\ 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+ \\ 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+ \\ 2^0+2^1+2^2+2^3+2^4+2^5+2^6+ \\ 2^0+2^1+2^2+2^3+2^4+2^5+ \\ 2^0+2^1+2^2+2^3+2^4+ \\ 2^0+2^1+2^2+2^3+ \\ 2^0+2^1+2^2+ \\ 2^0+2^1 \\ 2^0= \\ =2^{10}-1+2^9-1+2^8-1+2^7-1+2^6-1+2^5-1+2^4-1+2^3-1+2^2-1+2-1= \\ =2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}-10= \\ =2\frac{2^{10}-1}{2-1}-10=2^{11}-12=2048-12=2036 \end{array} $$ Thus: $$\sum_{k=1}^{10}{\dfrac{k}{2^k}}=\frac{2036}{1024}=\frac{509}{256}$$

KonKan
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2

$$ S=\sum_{k=1}^{10}{k\over2^k}=2\sum_{k=1}^{10}{k\over2^{k+1}}=2\sum_{k=2}^{11}{k-1\over 2^k}=2\sum_{k=1}^{10}{k-1\over 2^k}+{10\over2^{10}}=2S-\sum_{k=1}^{10}{1\over2^{k-1}}+{10\over2^{10}}\\ \implies S=2-{2\over2^{10}}-{10\over2^{10}}=2-{3\over2^8} $$

Jack's wasted life
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2

$$\begin{align} \sum_{k=1}^N \frac{k}{2^k}&=\sum_{k=1}^N \frac{1}{2^k}\sum_{j=1}^k(1)\\\\ &=\sum_{j=1}^N\sum_{k=j}^N\frac{1}{2^k}\\\\ &=\sum_{j=1}^N\left(\frac{(1/2)^j-(1/2)^{N+1}}{1-1/2}\right)\\\\ &=2\sum_{j=1}^N(1/2)^j-(1/2)^N\,N\\\\ &=2-(1/2)^{N-1}-(1/2)^N\,N \end{align}$$

For $N=10$, we have

$$\sum_{k=1}^{10} \frac{k}{2^k}=2-(1/2)^9-10(1/2)^{10}$$

Mark Viola
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0

Hint:

$$\sum_{k=1}^{10}\frac{k}{2^k} = \sum_{j = 1}^{10}\sum_{k=j}^{10}\frac{1}{2^k}$$ Apply power series formulae to ${\frac{1}{2^k}}$

runaround
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