Question

Find the sum to infinity of the series:
$1+\dfrac{2}{3}+\dfrac{6}{{{3}^{2}}}+\dfrac{10}{{{3}^{3}}}+\dfrac{14}{{{3}^{4}}}+........$

Answer 1

Hint: Let us suppose that the summation of the above series be “S” and equate the above series to “S” then multiply left hand and right hand side by $\dfrac{1}{3}$ then subtract this new series after multiplication by $\dfrac{1}{3}$ from the original series. After subtraction, you will find a new series which is a G.P. and we know the formula for the summation of a series which is in a G.P.

Complete step-by-step answer:
The infinite summation series given in the question is:
$1+\dfrac{2}{3}+\dfrac{6}{{{3}^{2}}}+\dfrac{10}{{{3}^{3}}}+\dfrac{14}{{{3}^{4}}}+........$
Let us assume that the summation the above series is “S” and equate the above series to “S”.
$S=1+\dfrac{2}{3}+\dfrac{6}{{{3}^{2}}}+\dfrac{10}{{{3}^{3}}}+\dfrac{14}{{{3}^{4}}}+........$
Multiplying the above equation by $\dfrac{1}{3}$ on both the sides of the equation we get,
$\begin{aligned} & S=1+\dfrac{2}{3}+\dfrac{6}{{{3}^{2}}}+\dfrac{10}{{{3}^{3}}}+\dfrac{14}{{{3}^{4}}}+........ \\\ & \Rightarrow \dfrac{1}{3}S=\dfrac{1}{3}+\dfrac{2}{{{3}^{2}}}+\dfrac{6}{{{3}^{3}}}+\dfrac{10}{{{3}^{4}}}+....... \\\ \end{aligned}$
Now, if you align the coefficient of the second series with the first series like if you write $\dfrac{1}{3}$ of the second series just below $\dfrac{2}{3}$ of the first series then subtraction of the two series will be quite easy. Similarly, you can write $\dfrac{2}{{{3}^{2}}}$ of the second series just below $\dfrac{6}{{{3}^{2}}}$ of the first series and you can write the other elements of the series and then subtract them. So, subtracting the above two series in a way we have just described we get,
$\begin{aligned} & \text{ }S=1+\dfrac{2}{3}+\dfrac{6}{{{3}^{2}}}+\dfrac{10}{{{3}^{3}}}+\dfrac{14}{{{3}^{4}}}+........ \\\ & \dfrac{-\dfrac{1}{3}S=\text{ }\dfrac{1}{3}+\dfrac{2}{{{3}^{2}}}+\dfrac{6}{{{3}^{3}}}+\dfrac{10}{{{3}^{4}}}+.......}{\dfrac{2}{3}S=1+\dfrac{1}{3}+\dfrac{4}{{{3}^{2}}}+\dfrac{4}{{{3}^{3}}}+\dfrac{4}{{{3}^{4}}}.......} \\\ \end{aligned}$
Simplifying the above equation we get,
$\begin{aligned} & \dfrac{2}{3}S=1+\dfrac{1}{3}+\dfrac{4}{{{3}^{2}}}+\dfrac{4}{{{3}^{3}}}+\dfrac{4}{{{3}^{4}}}+.............. \\\ & \Rightarrow \dfrac{2}{3}S=\dfrac{4}{3}+\dfrac{4}{{{3}^{2}}}+\dfrac{4}{{{3}^{3}}}+\dfrac{4}{{{3}^{4}}}+............ \\\ \end{aligned}$
As you can see that the numerator of the right hand side of the above equation has 4 in common so we can take the 4 out from all the fractions on the right hand side of the equation.
$\dfrac{2}{3}S=4\left( \dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{3}^{4}}}+............ \right)$……….. Eq. (1)
In the above equation, you can see that on the right hand side apart from 4 the series is in G.P.
$\left( \dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{3}^{4}}}+............ \right)$
The above infinite summation series is in G.P. with the first term is $\dfrac{1}{3}$ and the common ratio as $\dfrac{1}{3}$ so using the first term and common ratio in the formula for infinite summation of a G.P.
We know that the summation of an infinite G.P. is equal to:
${{S}_{\infty }}=\dfrac{a}{1-r}$
In the above formula, “a” is the first term of the series and “r” is the common ratio of the series. Plugging the values of first term and common ratio we get the summation of the infinite series.
${{S}_{\infty }}=\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}$
$\Rightarrow {{S}_{\infty }}=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}=\dfrac{1}{2}$
Substituting the above value of summation of infinite series in eq. (1) we get,
$\begin{aligned} & \dfrac{2}{3}S=4\left( \dfrac{1}{3}+\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{3}}}+\dfrac{1}{{{3}^{4}}}+............ \right) \\\ & \Rightarrow \dfrac{2}{3}S=4\left( \dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{2}{3}S=2 \\\ & \Rightarrow S=3 \\\ \end{aligned}$
From the above solution, we have found the summation of the infinite series as 3.

Note: The point to be noted that the formula for infinite G.P. that we have applied above is for the common ratio between -1 to 1. And in the given series the common ratio is $\dfrac{1}{3}$ which lies between -1 to 1.