Question

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+....\frac{1}{n(n+1)}$ equals
[AMU 1995; RPET 1996; UPSEAT 1999, 2001]

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

Answer 1

The correct option is B $\frac{n}{n+1}$

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