Question

The sum of the infinite series $1 + \left( {\dfrac{2}{3}} \right) + \left( {\dfrac{7}{{{3^2}}}} \right) + \left( {\dfrac{{12}}{{{3^3}}}} \right) + \left( {\dfrac{{17}}{{{3^4}}}} \right) + \left( {\dfrac{{22}}{{{3^5}}}} \right) + ........$ is equal to
1. $\dfrac{9}{4}$
2. $\dfrac{{15}}{4}$
3. $\dfrac{{13}}{4}$
4. $\dfrac{{11}}{4}$

Answer 1

Hint: In this question, an infinity series $1 + \left( {\dfrac{2}{3}} \right) + \left( {\dfrac{7}{{{3^2}}}} \right) + \left( {\dfrac{{12}}{{{3^3}}}} \right) + \left( {\dfrac{{17}}{{{3^4}}}} \right) + \left( {\dfrac{{22}}{{{3^5}}}} \right) + ........$ is given. First, divide the series by $3$then subtract both the equations. Solve further, while solving you’ll get the infinite G.P. Series and use the formula of sum of infinite terms of G.P..

Formula used:
Sum of infinite terms of G.P. –
$S = a + ar + a{r^2} + a{r^3} + - - - - - + \infty $
$S = \dfrac{a}{{1 - r}}$

Complete step by step solution: 
Given that,
$S = 1 + \left( {\dfrac{2}{3}} \right) + \left( {\dfrac{7}{{{3^2}}}} \right) + \left( {\dfrac{{12}}{{{3^3}}}} \right) + \left( {\dfrac{{17}}{{{3^4}}}} \right) + \left( {\dfrac{{22}}{{{3^5}}}} \right) + ........ + \infty - - - - - \left( 1 \right)$
Divide above equation by $3$,
$\dfrac{S}{3} = \left( {\dfrac{1}{3}} \right) + \left( {\dfrac{2}{{{3^2}}}} \right) + \left( {\dfrac{7}{{{3^3}}}} \right) + \left( {\dfrac{{12}}{{{3^4}}}} \right) + \left( {\dfrac{{17}}{{{3^5}}}} \right) + \left( {\dfrac{{22}}{{{3^6}}}} \right) + ........ + \infty - - - - - \left( 2 \right)$
Subtract equation (1) and (2)
$\dfrac{{2S}}{3} = 1 + \dfrac{1}{3} + \dfrac{5}{{{3^2}}} + \dfrac{5}{{{3^3}}} + - - - - - - + \infty $
$\dfrac{{2S}}{3} = 1 + \dfrac{1}{3} + \dfrac{5}{3}\left( {\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + - - - - - - + \infty } \right)$
$\dfrac{{2S}}{3} = 1 + \dfrac{1}{3} + \dfrac{5}{3}\left( {\dfrac{{\left( {\dfrac{1}{3}} \right)}}{{1 - \left( {\dfrac{1}{3}} \right)}}} \right)$
$\dfrac{{2S}}{3} = 1 + \dfrac{1}{3} + \dfrac{5}{3}\left( {\dfrac{1}{3} \times \dfrac{3}{2}} \right)$
$\dfrac{{2S}}{3} = 1 + \dfrac{1}{3} + \dfrac{5}{6}$
$\dfrac{{2S}}{3} = \dfrac{{13}}{6}$
$S = \dfrac{{13}}{4}$
Hence, Option (3) is the correct answer.

Note: The key concept involved in solving this problem is the good knowledge of Series and sequence. Students must know whether the series is in A.P. or G.P.. For infinite and finite series formulas are changed. So, apply carefully.