Question

$n(n+1)(n+4)$.

Answer 1

$a_n=n(n+1)(n+4)=n(n^2+5n+4)=n^3+{5n}^2+4n\therefore S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n{k}^3+5\sum _{k=1}^n{k}^2+4\sum _{k=1}^{n}k=\frac{n^2{(n+1)}^2}{4}+\frac{5n(n+1)2n+1)}{6}+\frac{4n(n+1)}{2}=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4]=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+20n+10+24}{6}\right]=\frac{n(n+1)}{2}\left[\frac{3n^2+23n+34}{6}\right]=\frac{n(n+1)(3^2+23n+34)}{12}$