$a)S\left(n\right)=\frac{1}{3}{n}^3+\frac{1}{2}{n}^2+\frac{1}{6}n=\frac{1}{6}n\left(2n^2+3n+1\right)=\ast$

$\Delta =3^2-4\cdot 2\cdot 1=9-8=1$

$\sqrt{\Delta }=1$

$n_1=\frac{-3-1}{2\cdot 2}=-\frac{4}{4}=-1$

$n_2=\frac{-3+1}{2\cdot 2}=-\frac{2}{4}=-\frac{1}{2}$

$\ast =\frac{1}{6}n\cdot 2\left(n+1\right)\left(n+\frac{1}{2}\right)=\frac{1}{3}n\left(n+1\right)\left(n+\frac{1}{2}\right)$