Question

The sum to infinity of the series $1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....$ is:
$\left( a \right)2$
$\left( b \right)3$
$\left( c \right)4$
$\left( d \right)6$

Answer 1

Hint: In this particular question first find out which trend is following the given series i.e. whether it is following the arithmetic series, geometric series, or harmonic series after identifying use directly the formula of the sum of an infinite series so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given series
$1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....$
Let, $S = 1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....$.................... (1)
Now divide by 3 throughout in equation (1) we have,
$ \Rightarrow \dfrac{S}{3} = \dfrac{1}{3} + \dfrac{2}{{{3^2}}} + \dfrac{6}{{{3^3}}} + \dfrac{{10}}{{{3^4}}} + \dfrac{{14}}{{{3^5}}} + ....$................. (2)
Now subtract equation (2) from equation (1) we have,
$ \Rightarrow S - \dfrac{S}{3} = \left( {1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....} \right) - \left( {\dfrac{1}{3} + \dfrac{2}{{{3^2}}} + \dfrac{6}{{{3^3}}} + \dfrac{{10}}{{{3^4}}} + \dfrac{{14}}{{{3^5}}} + ....} \right)$
$ \Rightarrow \dfrac{{2S}}{3} = \left( 1 \right) + \left( {\dfrac{2}{3} - \dfrac{1}{3}} \right) + \left( {\dfrac{6}{{{3^2}}} - \dfrac{2}{{{3^2}}}} \right) + \left( {\dfrac{{10}}{{{3^3}}} - \dfrac{6}{{{3^3}}}} \right) + \left( {\dfrac{{14}}{{{3^4}}} - \dfrac{{10}}{{{3^4}}}} \right) + ........\infty $
Now simplify it we have,
$ \Rightarrow \dfrac{{2S}}{3} = \left( 1 \right) + \left( {\dfrac{1}{3}} \right) + \left( {\dfrac{4}{{{3^2}}}} \right) + \left( {\dfrac{4}{{{3^3}}}} \right) + \left( {\dfrac{4}{{{3^4}}}} \right) + ........\infty $
$ \Rightarrow \dfrac{{2S}}{3} = \left( {\dfrac{4}{3}} \right) + \left( {\dfrac{4}{{{3^2}}}} \right) + \left( {\dfrac{4}{{{3^3}}}} \right) + \left( {\dfrac{4}{{{3^4}}}} \right) + ........\infty $
$ \Rightarrow S = \dfrac{3}{2}\left[ {\left( {\dfrac{4}{3}} \right) + \left( {\dfrac{4}{{{3^2}}}} \right) + \left( {\dfrac{4}{{{3^3}}}} \right) + \left( {\dfrac{4}{{{3^4}}}} \right) + ........\infty } \right]$
Now take $\dfrac{4}{3}$ common we have,
\[ \Rightarrow S = \dfrac{3}{2} \times \dfrac{4}{3}\left[ {1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ........\infty } \right]\]
\[ \Rightarrow S = 2\left[ {1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ........\infty } \right]\]
So the common ratio of the above series is $\dfrac{{\dfrac{1}{3}}}{1} = \dfrac{1}{3},\dfrac{{\dfrac{1}{{{3^2}}}}}{{\dfrac{1}{3}}} = \dfrac{1}{3}$
So, as we see that the above series follow the trend of geometric progression with a constant common ratio.
Where the first term of the series is, a = 1, common ratio, r = $\dfrac{1}{3}$.
Now as we know that the formula of the sum of the infinite series of a G.P is given as,
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$, where a = first term and d = common ratio.
Now substitute the values in the above equation we have,
$ \Rightarrow {S_\infty } = 2\left( {\dfrac{1}{{1 - \dfrac{1}{3}}}} \right) = 2\left( {\dfrac{1}{{\dfrac{{3 - 1}}{3}}}} \right) = 2\left( {\dfrac{3}{2}} \right) = 3$
So this is the required sum to infinity of the given series.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the sum of an infinite G.P series which is stated above, so simplify substitute the values of the first term and common ratio in the formula and simplify as above we will get the required answer. Here the given series is not in G.P so after doing some operations on the given series we convert this into a standard form of G.P and then find a common ratio.