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

1. $2$
2. $3$
3. $4$
4. $6$

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

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

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

$\therefore S=3$
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edited