$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$ is equal to

  • A
    $\frac{n}{6n-4}$
  • B
    $\frac{n}{6n+3}$
  • C
    $\frac{n}{6n+4}$
  • D
    $\frac{n+1}{6n+4}$

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If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to

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