For any $n \in N$,$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \ldots + \frac{1}{(3n-1)(3n+2)} = $

  • A
    $\frac{n}{6n+4}$
  • B
    $\frac{n^2}{6n+4}$
  • C
    $\frac{1}{2} \cdot \frac{n^2}{6n+4}$
  • D
    $\frac{n}{6n^2+4}$

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