Question

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

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

Answer 1

Answer: (E)

$n+\frac{1}{2}(1-3^{-n})$

Given, $n$ series is
$\frac{4}{3},\frac{10}{9},\frac{28}{27},\frac{82}{81},\frac{244}{243},...$
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
$S_n=n+\frac{1}{2}(1-3^{-n})$
Put $n=2$
$S_2=2+\frac{1}{2}(1-\frac{1}{3^2})$
$=2+\frac{1}{2}\left(\frac{8}{9}\right)=2+\frac{4}{9}$
$=\frac{22}{9},$ which is true.
Hence, option E is correct.