Q117 / 170 · Manipal (MET) 2015 · Mathematics
The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots$ is
- A$2$
- Correct answerB$3$
- C$4$
- D$6$
Solution
Let $S$ be the sum of the given series.
i.e.
$S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\ldots$
$\Rightarrow \quad(S-1)=\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\ldots$ ...(i)
$\Rightarrow \quad(S-1) \times \frac{1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\ldots$ ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$\frac{2}{3}(S-1)=\frac{2}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots$
$\Rightarrow \frac{2}{3}(S-1)=\frac{2}{3}+\frac{\frac{4}{3^2}}{1-\frac{1}{3}}$
$\Rightarrow \frac{2}{3}(S-1)=\frac{2}{3}+\frac{2}{3}$
$\Rightarrow S-1 \Rightarrow S=3$
Hence, the correct answer is option: 'B'.