Question

Sum of $\frac{1}{3},\frac{1}{15},\frac{1}{35},\frac{1}{63},\dots \dots 2n$ terms is

• A. $\frac{1}{2n+1}$
• B. $\frac{1}{n+1}$
• C. $\frac{2n}{4n+1}$
• D. $\frac{1}{2n-1}$

Answer 1

Answer: (C)

$\frac{2n}{4n+1}$

$t_1=\frac{1}{3}=\frac{1}{1\times 3};t_2=\frac{1}{15}=\frac{1}{3\times 5};t_3=\frac{1}{35}=\frac{1}{5\times 7}$

$t_n=\frac{1}{(2n-1)(2n+1)}$

$\therefore$ Sum of $Zn$ terms $={\sum }_{n=1}^{2n}{t}_n$

$={\sum }_{n=1}^{2n}\frac{1}{(2n-1)(2n+1)}$

$=\frac{1}{2}{\sum }_{n=1}^{2n}\frac{1}{2n-1}-\frac{1}{2n+1}$

$=\frac{1}{2}[1-\frac{1}{4n+1}]=\frac{2n}{4n+1}$