ISI2018-MMA-24

Question

The sum of the infinite series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots $ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

$S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+….$
$3S=5+\frac{6}{3}+\frac{10}{3^2}+\frac{14}{3^3}+….$

$3S-S=4+\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+...$

$2S=4\left (\frac{1}{1-\frac{1}{3}} \right )=4 \times \frac{3}{2}=6$

$\therefore S=3$