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

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(D) Given $t_{n} = \frac{1}{4}(n+2)(n+3)$.We need to find $S = \sum_{n=1}^{2003} \frac{1}{t_{n}} = \sum_{n=1}^{2003} \frac{4}{(n+2)(n+3)}$.Using the m

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$\frac{4006}{3009}$

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(D) Given $t_{n} = \frac{1}{4}(n+2)(n+3)$.We need to find $S = \sum_{n=1}^{2003} \frac{1}{t_{n}} = \sum_{n=1}^{2003} \frac{4}{(n+2)(n+3)}$.Using the method of partial fractions,$\frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$.Thus,$S = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} ight)$.This is a telescoping series:$S = 4 \left[ (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \dots + (\frac{1}{2005} - \frac{1}{2006}) ight]$.$S = 4 \left( \frac{1}{3} - \frac{1}{2006} ight) = 4 \left( \frac{2006 - 3}{3 	imes 2006} ight) = 4 \left( \frac{2003}{6018} ight) = \frac{8012}{6018} = \frac{4006}{3009}$.

If $t_{n} = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \dots$,then find the value of $\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \dots + \frac{1}{t_{2003}}$.

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