Sum the series

Question

1*2+2*3+3*4+ ... +2019*2020

Answer 1

General term = n(n+1) for n ∈ N
We want:
∑ n(n+1) for n=1 to 2019
= ∑ n^2 + n
= ∑ n^2 + ∑ n
= n(n+1)(2n+1)/6 + n(n+1)/2 , replace n with 2019 and evaluate