We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
179
2
avatar+1114 

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+5097 
+2

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

.
 Sep 5, 2018
 #1
avatar+5097 
+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+22196 
+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

5 Online Users

avatar