Question

Value of $\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\dots$ to ${}^{\prime }{n}^{\prime }$ terms is equal to:

• A. $\frac{4n^2+8n}{12(2n+1)(2n+3)}$
• B. $\frac{1}{4}(\frac{1}{(2n+1)(2n+3)}-\frac{1}{3})$
• C. $\frac{4n^2+6n}{12(2n+1)(2n+3)}$
• D. None of these

Answer 1

The correct option is A $\frac{4n^2+8n}{12(2n+1)(2n+3)}$

$\frac{1}{1\cdot 3\cdot 5}+\frac{1}{3\cdot 5\cdot 7}+\cdots$ General term of the given sequence is : $T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}$

$=\frac{1}{4}(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}$

$\frac{1}{3/5}\cdot \frac{1}{5\cdot 7}----$

$\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)})$

= $\frac{1}{4}(\frac{1}{3}-\frac{1}{(2n+1)(2n+3)})$

=$\frac{1}{12}\frac{4n^2+8n}{(2n+1)(2n+3)}$