First slide

Special Series in Sequences and Series

Question

$If t_{n} = \frac{1}{4} (n + 2) (n + 3) for n = 1, \underline{\underline{2, 3}, \dots} then \frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \dots + \frac{1}{t_{2003}} is$

Moderate

A

$\frac{4006}{3006}$
B

$\frac{4003}{3007}$
C

$\frac{4006}{3008}$
D

$\frac{4006}{3009}$

Solution

$\begin{array}{l} \frac{1}{t_{n}} = \frac{4}{(n + 2) (n + 3)} = 4 (\frac{1}{n + 2} - \frac{1}{n + 3}) \\ \begin{array}{l} \sum_{n = 1}^{2003} \frac{1}{t_{n}} = 4 [(\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \dots + (\frac{1}{2005} - \frac{1}{2006})] = 4 [\frac{1}{3} - \frac{1}{2006}] \\ = \frac{2003 \times 4}{3 \times 2006} = \frac{4006}{3009} \end{array} \end{array}$

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