Question

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

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The correct option is B 3

$s=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+....s-1=\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+...\frac{s-1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+...(s-1)-\left(\frac{s-1}{3}\right)=\frac{2}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+......(s-1)-\left(\frac{s-1}{3}\right)=\frac{2}{3}+\frac{4}{3^2}[1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....]\frac{3s-3-s+1}{3}=\frac{2}{3}+\frac{4}{3^2}(\frac{1}{1}-\frac{1}{3})\frac{2}{3}(s-1)=2\times \frac{2}{3}s-1=2s=3$