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

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Correct option is D

1tn=4(n+2)(n+3)=41n+2−1n+3∑n=12003 1tn=413−14+14−15+…+12005−12006=413-12006=2003×43×2006=40063009

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If tn=14(n+2)(n+3) for n=1,2,3_,…_ then 1t1+1t2+1t3+…+1t2003 is

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