Question

If $t_{n} = \frac{1}{4} \left(n+2\right)\left(n+3\right)$ for n = 1,2,3... then $ \frac{1}{t_{1} } + \frac{1}{t_{2}} + \frac{1}{t_{3}} +....+ \frac{1}{t_{2005}} = $

• A. $\frac{4004}{3005}$
• B. $\frac{3005}{4004}$
• C. $\frac{8020}{6024}$
• D. $\frac{8008}{6015}$

Answer 1

Answer: (C)

$\frac{1}{t_{n}} = \frac{4}{\left(n+2\right)\left(n+3\right)} = \frac{4}{n+2}-\frac{4}{n+3}$
Now $ \left(\frac{4}{3} - \frac{4}{4}\right)+\left(\frac{4}{4}-\frac{4}{5}\right)+\left(\frac{4}{5}- \frac{4}{6}\right) +....$
$+ \left(\frac{4}{2007} - \frac{4}{2008}\right) =$
$ \frac{4}{3} - \frac{4}{2008}= \frac{8020}{6024} $