A) \[\frac{4006}{3006}\]
B) \[\frac{4003}{3007}\]
C) \[\frac{4006}{3008}\]
D) \[\frac{4006}{3009}\]
Correct Answer: D
Solution :
\[{{t}_{n}}=\frac{1}{4}(n+2)(n+3)\], then \[\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}+\frac{1}{{{t}_{3}}}+.....+\frac{1}{{{t}_{2003}}}\] \[=4\left[ \frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{(2005).(2006)} \right]\] \[=4\left[ \frac{1}{3}-\frac{1}{2006} \right]\]\[=4.\frac{2003\,}{3(2006)}=\frac{4006}{3009}\].
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