If $T_{n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1,2,3,\ldots,$ then $\frac{1}{T_{1}} + \frac{1}{T_{2}} + \ldots + \frac{1}{T_{2003}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Since, $T_{n} = \frac{1}{4}(n + 2)(n + 3)$
$\frac{1}{T_{n}} = \frac{4}{(n + 2)(n + 3)} = 4\left\lbrack \frac{1}{n + 2} - \frac{1}{n + 3} ightbrack$
$	herefore\frac{1}{T_{1}} + \frac{1}{T_{2}} + ... + \frac{1}{T_{2003}} = 4\left( \frac{1}{3} - \frac{1}{4} ight) + 4\left( \frac{1}{4} - \frac{1}{5} ight)$
$+ ... + 4\left( \frac{1}{2005} - \frac{1}{2006} ight)$
$= 4\left( \frac{1}{3} - \frac{1}{2006} ight) = \frac{4 	imes 2003}{3 	imes 2006} = \frac{4006}{3009}$

Q. If $T_{n} = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1,2,3,\ldots,$ then $\frac{1}{T_{1}} + \frac{1}{T_{2}} + \ldots + \frac{1}{T_{2003}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?