Question

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

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

Answer 1

The correct option is B

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

$Let{T}_n\text{be the sum of n terms of the given series.}Thus,wehave:$ $T_n=\frac{3^n-1}{3^n}=1-\frac{1}{3^n}Now,S_n={\sum }_{k=1}^n{T}_k={\sum }_{k=1}^n[1-\frac{1}{3^k}]={\sum }_{k=1}^{n}1-{\sum }_{k=1}^n\frac{1}{3^k}=n-[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}]=n-\frac{1}{3}\left[\frac{1-\left(\frac{1}{3^n}\right)}{1-\frac{1}{3}}\right]=n-\frac{1}{2}[1-{\left(\frac{1}{3}\right)}^n]=n-\frac{1}{2}[1-3^{-n}]$