Question

Find the sum to $n$ terms of the series $\frac{1}{1\cdot3} +\frac{1}{3\cdot5}+\frac{1}{5\cdot7} +...$

• A. $\frac{n}{\left(2n-1\right)}$
• B. $\frac{n}{\left(2n+1\right)}$
• C. $n+1$
• D. $n-1$

Answer 1

Answer: (B)

Let $T_{r}$ be the $r^{th}$ term of the given series. Then,
$ T_{r} = \frac{1}{ \left(2r-1\right)\left(2r+1\right)}, r= 1, 2, 3, ..., n $
$ \Rightarrow T_{r} = \frac{1}{2} \left(\frac{1}{2r-1} - \frac{1}{ 2r+1}\right) $
$\therefore $ Required sum
$ = \sum\limits_{r=1}^{n}T_{r} = \frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{2n-1} -\frac{1}{2n+1}\right)\right]$
$ = \frac{1}{2}\left[1-\frac{1}{2n+1}\right] = \frac{n}{2n+1}$