# Find the sum of $n$ terms of the series whose $n^{th}$ term is given by $n(n+1)(n+4)$
$\begin{array}{1 1}\large\frac{n(n+1)(3n^2+23n+34)}{24} \\ \large\frac{n(n+1)(3n^2+23n+24)}{12} \\ \large\frac{n(n+1)(3n^2+20n+34)}{12}\\\large\frac{n(n+1)(3n^2+23n+34)}{12}\end{array}$