Question

Sum of the series
${\sum }_{r=0}^n{(-1)}^r\left({{}^{n}C}_r\right)[\frac{2^r}{3^r}+\frac{8^r}{3^{2r}}+\frac{{26}^r}{3^{3r}}+....upto\infty ]$ is

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

Answer 1

The correct option is C $\frac{1}{3^n-1}$

$S={\sum }_{r=0}^n{(-1)}^r{C}_r{\left(\frac{2}{3}\right)}^r+{\sum }_{r=0}^n{(-1)}^r{C}_r{\left(\frac{8}{9}\right)}^r+....upto\infty$
$={\left(1-\frac{2}{3})\right)}^n+{(1-\frac{8}{9})}^n+....upto\infty$
$=\frac{1}{3^n}+\frac{1}{3^{2n}}+....=\frac{{\frac{1}{3}}^n}{1-{\frac{1}{3}}^n}=\frac{1}{3^n-1}$