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Evaluate the sum \(\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{k}{2^k} + \cdots \)

 Sep 4, 2018

Best Answer 

 #1
avatar+4436 
+2

\(\sum \limits_{k=1}^\infty~x^k = \dfrac{x}{1-x} \\ \\ \text{differentiate both sides} \\ \\ \dfrac {d}{dx} \sum \limits_{k=1}^\infty~x^k = \dfrac{d}{dx} \dfrac{x}{1-x} \\ \\ \sum \limits_{k=1}^\infty~k x^{k-1} = \dfrac{1}{(1-x)^2} \\ \\ \text{mutliply both sides by }x \\ \\ \sum \limits_{k=1}^\infty~k x^k = \dfrac{x}{(1-x)^2} \\ \\ \text{let }x=\dfrac 1 2 \\ \\ \sum \limits_{k=1}^\infty~\dfrac{k}{2^k} = \dfrac{\frac 1 2}{\left(1-\frac 1 2\right)^2} = 2\\ \\\)

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 Sep 5, 2018
 #1
avatar+4436 
+2
Best Answer

\(\sum \limits_{k=1}^\infty~x^k = \dfrac{x}{1-x} \\ \\ \text{differentiate both sides} \\ \\ \dfrac {d}{dx} \sum \limits_{k=1}^\infty~x^k = \dfrac{d}{dx} \dfrac{x}{1-x} \\ \\ \sum \limits_{k=1}^\infty~k x^{k-1} = \dfrac{1}{(1-x)^2} \\ \\ \text{mutliply both sides by }x \\ \\ \sum \limits_{k=1}^\infty~k x^k = \dfrac{x}{(1-x)^2} \\ \\ \text{let }x=\dfrac 1 2 \\ \\ \sum \limits_{k=1}^\infty~\dfrac{k}{2^k} = \dfrac{\frac 1 2}{\left(1-\frac 1 2\right)^2} = 2\\ \\\)

Rom Sep 5, 2018
 #2
avatar+21848 
+8

Evaluate the sum

\(\large{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \cdots + \dfrac{k}{2^k} + \cdots}\)
\large{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \cdots + \dfrac{k}{2^k} + \cdots}

 

Binomial power:

\(\begin{array}{|rcll|} \hline (1-x)^{-2} &=& \dbinom{-2}{0} - \dbinom{-2}{1}x + \dbinom{-2}{2}x^2- \dbinom{-2}{3}x^3+ \dbinom{-2}{4}x^4 -\dbinom{-2}{5}x^5 \\ && +- \cdots \infty \\\\ &=& 1 -(-2)x + \dfrac{(-2)(-2-1)}{2\cdot 1}x^2 - \dfrac{(-2)(-2-1)(-2-2)}{3\cdot 2\cdot 1}x^3 \\ && +\dfrac{(-2)(-2-1)(-2-2)(-2-3)}{4\cdot 3\cdot 2\cdot 1}x^4 \\ && -\dfrac{(-2)(-2-1)(-2-2)(-2-3)(-2-4)}{5\cdot 4\cdot 3\cdot 2\cdot 1}x^5 +- \cdots \infty \\\\ &=& 1 +2x + \dfrac{(-2)(-3)}{2}x^2 - \dfrac{(-2)(-3)(-4)}{3\cdot 2}x^3 \\ && +\dfrac{(-2)(-3)(-4)(-5)}{4\cdot 3\cdot 2}x^4 -\dfrac{(-2)(-3)(-4)(-5)(-6)}{5\cdot 4\cdot 3\cdot 2}x^5 +- \cdots \infty \\\\ &=& 1 +2x + 3x^2 +4x^3 + 5x^4+6x^5 + \cdots \infty \\\\ \hline \\ x\cdot (1-x)^{-2} &=& x\cdot \left( 1 +2x + 3x^2 +4x^3 + 5x^4+6x^5 + \cdots \infty \right) \\\\ &=& x +2x^2 + 3x^3 +4x^4 + 5x^5+6x^6 + \cdots + k x^k + \cdots \infty \quad | \quad x=\dfrac12 \\\\ \dfrac12\left(1-\dfrac12 \right)^{-2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac12\left(\dfrac12 \right)^{-2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac12\left(\dfrac21 \right)^{2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac42 &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ 2 &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\ \hline \end{array}\)

 

 

\(\mathbf{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty=2 } \\\)

 

 

laugh

 Sep 5, 2018
edited by heureka  Sep 5, 2018
edited by heureka  Sep 5, 2018

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