If t_n = frac{1}{4}(n+2)(n+3) for n = 1, 2, 3 | vedclass

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(D) We are given $t_n = \frac{1}{4}(n+2)(n+3)$.Thus,$\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.Using partial fractions,we can write $\frac{1}{t_n} = 4 \le

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$\frac{4006}{3009}$

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(D) We are given $t_n = \frac{1}{4}(n+2)(n+3)$.Thus,$\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.Using partial fractions,we can write $\frac{1}{t_n} = 4 \left[ \frac{1}{n+2} - \frac{1}{n+3} ight]$.Let $S = \sum_{n=1}^{2003} \frac{1}{t_n} = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} ight)$.This is a telescoping sum:$S = 4 \left[ \left( \frac{1}{3} - \frac{1}{4} ight) + \left( \frac{1}{4} - \frac{1}{5} ight) + \ldots + \left( \frac{1}{2005} - \frac{1}{2006} ight) ight]$.All intermediate terms cancel out,leaving:$S = 4 \left( \frac{1}{3} - \frac{1}{2006} ight)$.$S = 4 \left( \frac{2006 - 3}{3 	imes 2006} ight) = 4 \left( \frac{2003}{6018} ight) = \frac{8012}{6018} = \frac{4006}{3009}$.

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

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