Question

The sum of infinity of the series 2/3 – 4/3² + 2/3³ + 4/3⁴ is​

Answer 1

$\huge \bf \: answer$

Firstly let's rearrange it

$\sf\bigg(\dfrac{2}{3}+\dfrac{2}{3^3}+\dfrac{2}{3}^{4}.\bigg)-\bigg(\dfrac{4}{3}^{2}+\dfrac{4}{3}^{4}...\bigg)$

$\sf \: First \: series$

$2 \bigg(\dfrac{1}{3} + \dfrac{1}{ {3^{3} } + \dfrac {1}{3}^{5}...\bigg)$

It is a gp infinite series.

$\sf \: first \: term \: = \dfrac{1}{3}$

$\sf \: r \: = \dfrac{1}{ {3}^{3} } \div \dfrac{1}{3}$

$\sf \: r \: = \dfrac{1}{ {3}^{2} }$

Sum of infinite terms in gp

$\sf \: = \dfrac{a}{1 - r}$

$\sf \: = \dfrac{ \dfrac{1}{3} }{ \dfrac {1- }{3}^{2} }$

$\sf \: \dfrac{ \dfrac{1}{3} }{ \dfrac{8}{9}} = \dfrac {1}{3} \times \dfrac {9}{8}$

$\sf \bold { \dfrac{3}{8} }$

$\sf\: first \: series \: =$

$2( \dfrac{1}{3} + \dfrac{1}{ {3}^{3} } + \dfrac {1}{3}^{5}....) = 2( \dfrac{3}{8} </p><p>) = \frac{3}{8}$.

$\sf \: second \: series \:$

$\sf \: 4 ( \dfrac{1}{ {3}^{2} } - \dfrac{1}{ {3}^{4} }....)$

$\sf \: first \: term(a) = \dfrac {1}{3}^{2}$

$\sf \: r \: = \ffrac{1}{ {3}^{4} } \div \dfrac{1 }{ {3}^{2} }$

$\sf \: r \: = \dfrac{1}{ {3}^{2} }$

$\sf \: sum \: of \: infinite \: series \:$

$\sf \: \dfrac{a}{1 - r}$

$\frac{ \dfrac{1}{ {3}^{2} } }{1 - \dfrac{1}{3}^{2} }$

$\sf \: s \: = \dfrac{1}{9} \times \frac{9}{8}$

$\sf \: s \: = \dfrac{9}{8}$

$\sf \: Required \: sum \: series$

$\sf \: \sf \: 4 ( \dfrac{1}{ {3}^{2} } - \dfrac{1}{ {3}^{4} }....) = 4( \dfrac{1}{8} ) = \dfrac{1}{2}$

$\sf \: now$

$( \dfrac{2}{3} + \dfrac{2}{ {3}^{3} }) +\ffrac {2}{3}^{5} \: \sf \: - \: ( \dfrac{4}{ {3}^{2} }+ \dfrac{4}{ {3}^{4} } )$

$\sf \dfrac{3}{4} - \dfrac{1}{2}$

$\sf \: \dfrac{3}{4} - \dfrac{2}{4} = \dfrac{1}{4}$

Answer 2

Answer:

prefer to the attachment