If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

  • A
    $\frac{4006}{3006}$
  • B
    $\frac{4003}{3007}$
  • C
    $\frac{4006}{3008}$
  • D
    $\frac{4006}{3009}$

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