If \(t_n=\frac{1}{4}(n+2)(n+3)\) for n = 1, 2, 3,....., then \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots .+\frac{1}{t_{2003}}=\)

1
\( \frac{4006}{3006}\)
2
\( \frac{4003}{3007}\)
3
\( \frac{4006}{3008}\)
4
\( \frac{4006}{3009} \)

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