Question

The sum of the infinite series $\frac{1}{3}+\frac{3}{3.7}+\frac{5}{\mathrm{3.7.11}}+\frac{7}{\mathrm{3.7.11.15}}+.......$ is

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

Answer 1

Answer: (A)

The $n^{th}$ term of the series is
$t_n=\frac{2n-1}{\mathrm{3.7.11.}---\left(4n-1\right)}=\frac{1}{2}\left\{\frac{\left(4n-1\right)-1}{\mathrm{3.7.11.}---\left(4n-1\right)}\right\}$
$=\frac{1}{2}\left\{\frac{1}{\mathrm{3.7.11.}---\left(4n-5\right)}-\frac{1}{\mathrm{3.7.11.}---\left(4n-1\right)}\right\}$
$=\frac{1}{2}\left(u_{n-1}-u_n\right),n\ge 2$
Where, $u_n=\frac{1}{\mathrm{3.7.11.}---\left(4n-1\right)}$
$\Rightarrow S_n=t_1+\sum _{k=2}^n{t}_k=\frac{1}{3}+\frac{1}{2}\left(u_1-u_2+u_2-u_3---+u_{n-1}-u_n\right)$
$=\frac{1}{3}+\frac{1}{2}\left(u_1-u_n\right)=\frac{1}{3}+\frac{1}{2}\left(\frac{1}{3}-u_n\right)$
$\Rightarrow \underset{n\to \in fty}{\lim }{S}_n=\frac{1}{3}+\frac{1}{6}-\frac{1}{2}\underset{n\to \in fty}{\lim }{u}_n=\frac{1}{2}-0=\frac{1}{2}$