Question

Find the sum to $n$ terms of the series $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+...$

• 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
$=\sum _{r=1}^n{T}_r=\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right]$
$=\frac{1}{2}\left[1-\frac{1}{2n+1}\right]=\frac{n}{2n+1}$