Question

$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+..(n-3)$ terms

• A. $\frac{n}{n+2}$
• B. $\frac{n+1}{n(n+5)}$
• C. $\frac{n-3}{2n-5}$
• D. $\frac{(n-1)}{n(2n-3)}$

Answer 1

The correct option is C $\frac{n-3}{2n-5}$

Consider the $i^{th}$ term: $t_i=\frac{1}{(2i-1)(2i+1)}$.

We re-write it as :

$t_i=\frac{1}{2}\times \frac{(2i+1)-(2i-1)}{(2i-1)(2i+1)}=\frac{1}{2}\times (\frac{1}{2i-1}-\frac{1}{2i+1})$

Similarly, $t_{i+1}=\frac{1}{2}\times (\frac{1}{2i+1}-\frac{1}{2i+3})$

Observe that first time in $t_i$ matches with the second term of $t_{i-1}$, while second term of $t_i$ matches with the first term of $t_{i+1}$. This is an instance of telescoping summing. Only the first term of $t_1$ and the last term of $t_{n-3}$ will remain.

Answer = $\frac{1}{2}\times (1-\frac{1}{2n-5})$

$=\frac{1}{2}\times \left(\frac{2n-6}{2n-5}\right)$

$=\frac{1}{2}\times \left(\frac{2(n-3)}{2n-5}\right)$

$=\frac{n-3}{2n-5}$