Cubes Are King

I'll rather generalise this:
Expression is of the form:
$\Large\mathcal{S}=\color{#007fff}{\dfrac{\displaystyle \sum n^3}{\color{#D61F06}{(\displaystyle \sum n)\color{#20A900}{(\displaystyle \sum n^2)}}}}$ $\Large =\color{#007fff}{\dfrac{(\frac{n(n+1)}{2})^{^{^2}}}{\color{#D61F06}{(\frac{n(n+1)}{2})\color{#20A900}{(\frac{n(n+1)(2n+1)}{6})}}}}$ $\Large =\dfrac{3}{2n+1}$ For $n=2016 ,~ \mathcal{S}=\dfrac{3}{4033}$ $\Large \therefore 4033+3=\boxed{\color{#302B94}{4036}}$

$\small\boxed{{\color{#624F41}{\text{Formula used}}: \color{#D61F06}{\displaystyle \sum n=\dfrac{n(n+1)}{2}\\ \color{#20A900}{\displaystyle \sum n^2=\dfrac{n(n+1)(2n+1)}{6}\\ \color{#007fff}{\displaystyle \sum n^3=(\dfrac{n(n+1)}{2})^{^2}}}}}}$

$\frac{3}{2n+1}$ is the general form Because we know that: \sum n^{3} =\(\frac{(n(n+1))^{2}}{4} ), \sum n^{2} =\(\frac{n(n+1)(2n+1)}{6} and \sum n=\(\frac{n(n+1)}{2}