Question

If $t_n=\frac{1}{4}(n+2)(n+3)$ for $n=1,2,\mathrm{3....}$ then $\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+.........+\frac{1}{t_{2003}}=$

• A. $\frac{4006}{3006}$
• B. $\frac{4003}{3007}$
• C. $\frac{4006}{3008}$
• D. $\frac{4006}{3009}$

Answer 1

The correct option is D $\frac{4006}{3009}$

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

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

$=4[\frac{1}{n+2}-\frac{1}{n+3}]$

Now sum $=\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_{2002}}+\frac{1}{t_{2003}}$

$=4[\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}]$

$=4[\frac{1}{3}-\frac{1}{2006}]$

$=4\left[\frac{2006-3}{3\left(2006\right)}\right]$

$=\frac{2\left(2003\right)}{3009}=\frac{4006}{3009}$