Question

The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+...$ to $n$ terms is

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

Answer 1

The correct option is C $n+\frac{1}{2}(3^{-n}-1)$

$\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+...$
$=(1-\frac{1}{3})+(1-\frac{1}{3^2})+(1-\frac{1}{3^3})...+(1-\frac{1}{3^n})$
$=(1+1+...............\text{n times})-(\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^n})$
$=n-\frac{\left[1/3\right(1/3{)}^n-1]}{1/3-1}=n+\frac{1}{2}(3^{-n}-1)$