The sum to the infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots…
The sum to the infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots \ldots$ is
-
2
-
3
-
4
-
6
Solution
Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots$.
$\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\ldots$
Dividing (1) \& (2)
$
\begin{aligned}
& S\left(1-\frac{1}{3}\right)=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots . \\
& \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\ldots . .\right) \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)=\frac{4}{3}+\frac{4}{3^2} \frac{3}{2}=\frac{4}{3}+\frac{2}{3}=\frac{6}{2} \Rightarrow \frac{2}{3} S=\frac{6}{3} \Rightarrow S=3
\end{aligned}
$
Asked in: JEE Main 2009
Practice more Sequences and Series questions on Aicharya