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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1
If Tn=14(n+2)(n+3){T_n} = \frac{1}{4}\left( {n + 2} \right)\left( {n + 3} \right) for n=1,2,3,…,n = 1,2,3, \ldots , then 1T1+1T2+…+1T2003\frac{1}{{{T_1}}} + \frac{1}{{{T_2}}} + \ldots + \frac{1}{{{T_{2003}}}} is equal to
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Correct option: D
40063009\frac{{4006}}{{3009}}
Option D (40063009\frac{{4006}}{{3009}}) is correct because the sum of the series is 40063009\frac{{4006}}{{3009}}.

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