If \[{{t}_{n}}=\frac{1}{4}(n+2)\,(n+3)\] for n = 1, 2, 3,?? then \[\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}+\frac{1}{{{t}_{3}}}+....+\frac{1}{{{t}_{2003}}}=\] [EAMCET 2003]

If \[{{t}_{n}}=\frac{1}{4}(n+2)\,(n+3)\] for n = 1, 2, 3,?? then \[\fr

1.	If \[{{t}_{n}}=\frac{1}{4}(n+2)\,(n+3)\] for n = 1, 2, 3,?? then \[\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}+\frac{1}{{{t}_{3}}}+....+\frac{1}{{{t}_{2003}}}=\] [EAMCET 2003]
A.	\[\frac{4006}{3006}\]
B.	\[\frac{4003}{3007}\]
C.	\[\frac{4006}{3008}\]
D.	\[\frac{4006}{3009}\]
Answer» E.

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If \[{{t}_{n}}=\frac{1}{4}(n+2)\,(n+3)\] for n = 1, 2, 3,?? then \[\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}+\frac{1}{{{t}_{3}}}+....+\frac{1}{{{t}_{2003}}}=\] [EAMCET 2003]

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