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
$\frac{4}{3},\frac{10}{9},\frac{28}{27},\frac{82}{81},\frac{244}{243},\dots$
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