Question

The sum of$n$ terms of the series $\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.\dots$ is

Answer 1

Find the required sum.

Given series $\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.\dots n\mathrm{terms}$

$$\begin{gathered} S_n=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.\dots n\mathrm{terms} \\ =\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+...n\mathrm{terms} \\ =\underset{ntimes}{\underbrace{1+1+1...+1}}+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....n\mathrm{terms}\right) \end{gathered}$$

As, $\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....n\right)$ forms G.P.

Here, $a=\frac{1}{3}$ and$r=\frac{1}{3}$

So, by using sum of $n$ term of G.P. formula, we get

$$\begin{gathered} \Rightarrow S_n=n+\frac{1}{3}\left(\frac{1-{\left(\frac{1}{3}\right)}^n}{1-{\displaystyle \frac{1}{3}}}\right) \\ \Rightarrow S_n=n+\frac{1}{3}\left(\frac{1-\frac{1}{3^n}}{{\displaystyle \frac{2}{3}}}\right) \\ \Rightarrow S_n=n+\frac{1}{3}\left(\frac{\frac{3^n-1}{3^n}}{{\displaystyle \frac{2}{3}}}\right) \\ \Rightarrow S_n=n+\frac{3^n-1}{2·3^n} \\ \Rightarrow S_n=\frac{2n·3^n+3^n-1}{2·3^n} \end{gathered}$$

Hence, the sum of$n$ terms of the series $\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.\dots$ is $\frac{2n·3^n+3^n-1}{2·3^n}$.