Question

The sum to infinity of the series
$$1+2\text{/}3+6\text{/}{3}^2+10\text{/}{3}^3+14\text{/}{3}^4+...\text{is}\text{.}$$

Answer 1

Hint: Here, we will use the concept of sum of the infinite series of geometry progression i.e. GP. The sum of infinite series of a GP is possible only under certain conditions, otherwise the series will diverge and the sum will be infinity.

Complete step by step solution: Let the given Equation be S.
$$\text{S}=1+2\text{/}3+6\text{/}{3}^2+10\text{/}{3}^2+14\text{/}{3}^2+...(\text{i)}$$
First we need to check whether it is an AP series or GP. Series.
Looking at the pattern of the series, the numerator is AP. With a common difference of 4 and denominator is in GP. With a common ratio 3.
To make it a perfect, we divide equation (i) on both sides by 3.
$$\begin{array}{c}{\displaystyle \frac{\text{S}}{3}}={\displaystyle \frac{\left(1+2\text{/}3+6\text{/}{3}^2+10\text{/}{3}^3+14\text{/}{3}^4+...\mathrm{\infty }\right)}{3}}\\ \\ {\displaystyle \frac{\text{S}}{3}}={\displaystyle \frac{1}{3}}+2\text{/}{3}^2+6\text{/}{3}^3+10\text{/}{3}^4+14\text{/}{3}^5+...\mathrm{\infty }(\text{ii})\end{array}$$

For getting the perfect pattern out of these equations.
We subtract eq. (ii) from eq. (i).
$$\begin{array}{c}\text{S}-\text{S/}3=\left(1+2\text{/}3+6\text{/}{3}^2+10\text{/}{3}^3+14\text{/}{3}^4+...\mathrm{\infty }\right)\\ \left(1\text{/}3+2\text{/}{3}^2+6\text{/}{3}^3+10\text{/}{3}^4+...\mathrm{\infty }\right)\\ \\ ={\displaystyle \frac{25}{3}}=1+\left(2\text{/3}-1\text{/3}\right)+\left(6\text{/}{\text{3}}^2-2\text{/}{\text{3}}^2\right)+\left(10\text{/}{\text{3}}^3-6\text{/}{\text{3}}^3\right)+...\mathrm{\infty }\\ ={\displaystyle \frac{25}{3}}=1+{\displaystyle \frac{1}{3}}+4\text{/}{3}^2+4\text{/}{3}^3+4\text{/}{3}^4+...\mathrm{\infty }\\ \text{S}={\displaystyle \frac{3}{2}}1+1\text{/}3+4\text{/}{3}^2+4\text{/}{3}^2+4\text{/}{3}^4+...\mathrm{\infty }\\ =\text{S}={\displaystyle \frac{3+1}{2}}+{\displaystyle \frac{2}{3}}+{\displaystyle \frac{2}{3^2}}+{\displaystyle \frac{2}{3^3}}+{\displaystyle \frac{2}{3^4}}+...\mathrm{\infty }\\ =\text{S}=2+2\text{/}3+2\text{/}{3}^2+2\text{/}{3}^3+2\text{/}{3}^4+...\mathrm{\infty }\\ =\text{S}=2\text{/}{3}^0+2\text{/}{3}^1+2\text{/}{3}^2+2\text{/}{3}^3+2\text{/}{3}^4...\mathrm{\infty }(\text{iii})\end{array}$$
The above equation i.e. eq. (iii) is in the GP. Of infinite terms
here the common ratio of the equation is
$$={\displaystyle \frac{2\text{/}{\text{3}}^1}{2\text{/}{\text{3}}^0}}={\displaystyle \frac{2\text{/3}}{2\text{/1}}}={\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{3}}$$
$$\therefore$$common ratio =$${\displaystyle \frac{1}{3}}$$;
(because in a GP. a, ar, ar2. Common ratio = $${\displaystyle \frac{ar}{a}}=r$$)
Now, sum of the infinite series is GP is
$$\text{S}\mathrm{\infty }={\displaystyle \frac{a}{a-r}};$$ where O< r < l and
a id the first term of the GP and r this common ratio of an infinite GP.
So, $$\text{S}={\displaystyle \frac{2}{1-1\text{/}3}}$$
$\Rightarrow \text{S}={\displaystyle \frac{2}{{\displaystyle \frac{2}{3}}}}$
$\Rightarrow \text{S}=2\times {\displaystyle \frac{3}{2}}$
$\Rightarrow \text{S}=3$

Sum of the infinite series is 3.

Note: In this type of question, we will check for AP and GP them accordingly to solve the equation and numbers the sum formula for infinite series that is very important. In higher studies we’ll come across many infinite series. Some of them will be convergent means sum will be finite and some of them will be divergent means sum will be infinite. The famous example is ${\displaystyle \frac{1}{x^{\text{p}}}}$. this series is convergent for $\text{p}>1$ and convergent for $\text{p}<1$.