jmc

We can write the sum as $$\begin{array}{rl}0\cdot 1^2+1\cdot 2^2+3\cdot 4^2+\cdots +19\cdot 20^2& =\sum _{n=1}^{20}(n-1)n^2\\ & =\sum _{n=1}^{20}(n^3-n^2)\\ & =\sum _{n=1}^{20}{n}^3-\sum _{n=1}^{20}{n}^2\\ & =\frac{20^2\cdot 21^2}{4}-\frac{20\cdot 21\cdot 41}{6}\\ & =20\cdot 21\cdot \left(\frac{20\cdot 21}{4}-\frac{41}{6}\right)\\ & =\boxed{41230}.\end{array}$$