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1) Find
\[\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \dotsb.\]

 

2) Evaluate the series .$\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots$

 

SUBMIT ONE QUESTION PER POST and put your LaTex in a LaTex box - Melody

 Feb 16, 2020
edited by Melody  Feb 16, 2020
 #1
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1 - ∑[4^-n, n, 1, ∞ ] = 1 / 3


2 - ∑[ (3n - 2) / (3^n), n, 1, ∞ ] = 5 / 4

 Feb 16, 2020
 #2
avatar+24430 
+2

1) Find
\(\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb\) .

 

Geometric progression: \( a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots\)


\(\begin{array}{|rcll|} \hline \mathbf{\text{Geometric progression: } a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots,\qquad a=\dfrac{1}{4},\ r=\dfrac{1}{4} } \\ \hline \dfrac{1}{4},\ \dfrac{1}{4}*\left(\dfrac{1}{4}\right)^1,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^2,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^3,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^4,\ \ldots \\ \hline \end{array}\)

 

 

Infinite geometric series, the sum formula: \(\dfrac{a}{1-r}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \dfrac{\frac{1}{4}}{1-\frac{1}{4}} \quad |\quad a=\dfrac{1}{4},\ r=\dfrac{1}{4} \\\\ &=& \dfrac{4}{4*3} \\\\ &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

 

\(\mathbf{\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb} = \mathbf{\dfrac{1}{3}}\)

 

laugh

 Feb 17, 2020
edited by heureka  Feb 17, 2020
edited by heureka  Feb 17, 2020
 #3
avatar+24430 
+2

2) Evaluate the series

\(\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots \).

 

\(\begin{array}{|rcll|} \hline (1+0*3)\left(1*\frac{1}{3}\right) +(1+1*3)\left(1*\frac{1}{3^2}\right) +(1+2*3)\left(1*\frac{1}{3^3}\right) +(1+3*3)\left(1*\frac{1}{3^4}\right) + \dotsb \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}3})+({\color{red}1}+2*{\color{orange}3})+({\color{red}1}+3*{\color{orange}3})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}3}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}\frac{1}{3} } +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^1 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^2 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^3 +\dotsb +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\ \hline \end{array}\)

 

Formula:
sum of a infinite arithmetico-geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}\frac{1}{3}} }{1-{ \color{green}\frac{1}{3}} }\right) \Bigg( {\color{red}1} + { \color{orange}3} \left( \dfrac{ {\color{green}\frac{1}{3} }}{1-{\color{green}\frac{1}{3}} } \right) \Bigg) \\\\ s &=& \frac{1}{3}*\frac{3}{2} \Bigg( {\color{red}1} + \dfrac{1}{1-{\color{green}\frac{1}{3}} } \Bigg) \\\\ s &=& \frac{1}{2} *\frac{5}{2} \\\\ \mathbf{s} &=& \mathbf{\frac{5}{4}} \\ \hline \end{array}\)

 

\(\mathbf{\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots} = \mathbf{\dfrac{5}{4}}\)

 

laugh

 Feb 17, 2020

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