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Compute 1*1/2+2*1/3+3*1/8+...+n*1/2^n+...

 

if you guys could solve this then that would be great!

 Sep 13, 2018
 #1
avatar+21848 
+8

I assume: Compute 1*1/2+2*1/4+3*1/8+...+n*1/2^n+...

 

Compute:

\(\displaystyle s = 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty\)

 

\(\begin{array}{|lrll|} \hline &s =& \displaystyle 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty \\\\ &s =& \displaystyle 1\cdot \left(\dfrac12\right)^1+2\cdot\left(\dfrac12\right)^2+ 3\cdot \left(\dfrac12\right)^3+ 4\cdot \left(\dfrac{1}{2}\right)^4+\ldots + n\cdot \left(\dfrac{1}{2} \right)^n+\ldots \infty \\\\ && \displaystyle\qquad \text{We substitute:} \quad \boxed{x=\dfrac12} \\\\ &s =& \displaystyle 1x^1+2x^2+ 3x^3+ 4x^4+\ldots + nx^n+\ldots \infty \quad | \quad : x \\\\ &\dfrac{s}{x} =& \displaystyle 1x^0+2x^1+ 3x^2+ 4x^3+\ldots + nx^{n-1}+\ldots \infty \\\\ &\dfrac{s}{x} =& \displaystyle \sum \limits_{n=1}^{\infty} nx^{n-1} \quad | \quad \cdot \ dx \\\\ &\dfrac{s}{x}\ dx =& \displaystyle \sum \limits_{n=1}^{\infty} \left( nx^{n-1}\ dx \right) \quad | \quad \text{integrate both sides} \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \int \left(\sum \limits_{n=1}^{\infty} \left( nx^{n-1}\ dx \right) \right) \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \sum \limits_{n=1}^{\infty} x^{n} \quad | \quad \text{This is the sum of a infinite geometric series},\ a_1=x,\ r=x \\\\ && \displaystyle\qquad \text{The sum of this infinite geometric series is:} \quad \boxed{= \dfrac{x}{1-x},\ \quad |x| \lt 1 } \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \dfrac{x}{1-x} \quad | \quad \text{derivate both sides} \\\\ & \dfrac{s}{x} =& \displaystyle 1\cdot(1-x)^{-1}+x\cdot(-1)\cdot(1-x)^{-2}\cdot(-1) \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1-x+x}{(1-x)^2} \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1}{(1-x)^2} \\\\ & \mathbf{s=}& \mathbf{\displaystyle \dfrac{x}{(1-x)^2}} \quad | \quad x=\dfrac12 \\\\ & s= & \displaystyle \dfrac{\dfrac12}{\left(1-\dfrac12 \right)^2} \\\\ & s= & \displaystyle \dfrac{\dfrac12}{\left(\dfrac12 \right)^2} \\\\ & s= &\displaystyle \dfrac{1}{ \dfrac12 }\\\\ & \mathbf{s=}& \mathbf{\displaystyle 2} \\ \hline \end{array}\)

 

\(\displaystyle 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty = 2\)

 

laugh

 Sep 13, 2018
edited by heureka  Sep 13, 2018
 #3
avatar
+2

Thanks!

Guest Sep 13, 2018
 #2
avatar+27544 
+3

Heureka beat me to it! However, Here's my attempt anyway (like heureka, I've assumed the second term was written incorrectly in the original question):

 

.

 Sep 13, 2018

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