Big Computation

The sum can be rewritten as

$(2 \times 2^{2} + 3 \times 3^{2} + 4 \times 4^{2} + .... + 20 \times 20^{2}) - (2^{2} + 3^{2} + 4^{2} + ... + 20^{2}) =$

$\displaystyle\sum_{n=2}^{20} n^{3} - \sum_{n=2}^{20} n^{2} = \left(\left(\dfrac{20 \times 21}{2}\right)^{2} - 1\right) - \left(\dfrac{20 \times 21 \times 41}{6} - 1\right) = 210^{2} - 70 \times 41 = \boxed{41230}$ .

Summation (n(n+1)^2)
= S(n(n^2+2n+1))
= S(n^3+2n^2+n)
= S(n^3) + 2S(n^2) + S(n)
= n(n+1)/2 * [n(n+1)/2 + 2(2n+1)/3 + 1]
= n(n+1)/2 * [(3n^2+11n+10)/6]
= n(n+1)(n+2)(3n+5)/12

Plugging in n=19, we get
S = 41230