Correct Answer: $\frac{5}{16}$

Solution : This series is in the form of $\frac{1}{(3n-2)(3n+1)}$ where n starts from 1.
The difference between the two fractions:
$\frac{1}{(3n-2)(3n+1)} = \frac{1}{3} \left(\frac{1}{3n-2} - \frac{1}{3n+1}\right)$
The sum of the first five terms:
$⇒S_5 = \frac{1}{3} \left[\left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \left(\frac{1}{10} - \frac{1}{13}\right) + \left(\frac{1}{13} - \frac{1}{16}\right)\right]$
$⇒S_5 = \frac{1}{3} \left( 1 - \frac{1}{16}\right) = \frac{15}{48} = \frac{5}{16}$
Hence, the correct answer is $\frac{5}{16}$.