Question

The sum of first $n$ terms of the series
$\frac{4}{3}, \frac{10}{9}, \frac{28}{27}, \frac{82}{81}, \frac{244}{243}, \cdots$ is

• A. $n+\frac{1}{2} \left(1+3^{-n}\right)$
• B. $n-\frac{1}{2} \left(1+3^{-n}\right)$
• C. $n+\frac{1}{2} \left(2+3^{-n}\right)$
• D. $n+\frac{1}{2} \left(2-3^{-n}\right)$
• E. $n+\frac{1}{2} \left(1-3^{-n}\right)$

Answer 1

Answer: (E)

Given, $n$ series is
$\frac{4}{3}, \frac{10}{9}, \frac{28}{27}, \frac{82}{81}, \frac{244}{243}, \ldots $
Here, $S_{1}=\frac{4}{3}, S_{2}=\frac{4}{3}+\frac{10}{9}=\frac{12+10}{9}=\frac{22}{9}$
Now, taking option (e)
$S_{n}=n+\frac{1}{2}\left(1-3^{-n}\right) $
Put $n =1 $
$S_{1} =1+\frac{1}{2}\left(1-\frac{1}{3}\right)=1+\frac{1}{2}\left(\frac{2}{3}\right)$
$=\frac{4}{3}$, which is true.
Put $n=2$
$S_{2} =2+\frac{1}{2}\left(1-\frac{1}{3^{2}}\right) $
$=2+\frac{1}{2}\left(\frac{8}{9}\right)=2+\frac{4}{9}$
$=\frac{22}{9}$, which is true