An algebra problem by أحمد الحلاق

$\begin{aligned} S_n & = 1\times 2^2 + 2\times 3^2 + 3\times 4^2 +...+ n\times (n+1)^2 \\ & = \sum_{k=1}^n k(k+1)^2 \\ & = \sum_{k=1}^n (k^3+2k ^2 +k) \\ & = \frac {n^2(n+1)^2}4 + \frac {n(n+1)(2n+1)}3 + \frac {n(n+1)}2 \\ & = \frac {n(n+1)(3n^2+3n +8n+4 +6)}{12} \\ & = \frac {n(n+1)(3n^2+11n +10)}{12} \\ & = \frac {n(n+1)(n+2)(3n+5)}{12} \\ \implies S_{19} & = \frac {19\times 20 \times 21 \times 62}{12} = \boxed{41230} \end{aligned}$

Can also use sum[(k-1)*k^2] from 1 to 20 which gives sum[k^3-k^2].