If $ \sum^5_{n=1} \large\frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}$, then $k$ to equal to :

Question

$\begin{array}{1 1} \frac{55}{336} \\ \frac{17}{105} \\ \frac{1}{6} \\ \frac{19}{112} \end{array} $