1. [5] Compute
$$
1 \cdot 2^{2}+2 \cdot 3^{2}+3 \cdot 4^{2}+\cdots+19 \cdot 20^{2} .
$$
Answer: 41230 y Solution: We can write this as $\left(1^{3}+2^{3}+\cdots+20^{3}\right)-\left(1^{2}+2^{2}+\cdots+20^{2}\right)$, which is equal to $44100-2870=41230$.