Question

The sum of the infinite series $\frac{1}{3}+\frac{3}{3.7}+\frac{5}{3.7 . 11}+\frac{7}{3.7 . 11.15}+.......$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

The $n^{t h}$ term of the series is
$t_{n}=\frac{2 n - 1}{3.7 . 11 . - - - \left(4 n - 1\right)}=\frac{1}{2}\left\{\frac{\left(4 n - 1\right) - 1}{3.7 . 11 . - - - \left(4 n - 1\right)}\right\}$
$=\frac{1}{2}\left\{\frac{1}{3.7 . 11 . - - - \left(4 n - 5\right)} - \frac{1}{3.7 . 11 . - - - \left(4 n - 1\right)}\right\}$
$=\frac{1}{2}\left(u_{n - 1} - u_{n}\right), \, n\geq 2$
Where, $u_{n}=\frac{1}{3.7 . 11 . - - - \left(4 n - 1\right)}$
$\Rightarrow S_{n}=t_{1}+\displaystyle \sum _{k = 2}^{n} t_{k}=\frac{1}{3}+\frac{1}{2}\left(u_{1} - u_{2} + u_{2} - u_{3} - - - + u_{n - 1} - u_{n}\right)$
$=\frac{1}{3}+\frac{1}{2}\left(u_{1} - u_{n}\right)=\frac{1}{3}+\frac{1}{2}\left(\frac{1}{3} - u_{n}\right)$
$\Rightarrow \underset{n \rightarrow \in fty}{l i m}S_{n}=\frac{1}{3}+\frac{1}{6}-\frac{1}{2}\underset{n \rightarrow \in fty}{l i m}u_{n}=\frac{1}{2}-0=\frac{1}{2}$