Question

(2020^2−2018^2)(2020^2−2017^2)...(2020^2−0^2)/(2019^2−2018^2)(2019^2−2017^2)...(2019^2−0^2)

Answer 1

each bracket consists of a "difference of squares, so we have
(2)(4038)(3)(4037)(4)(4036)..(2020)(2020)/[(1)(4037)(2)(4036)(3)(4035)..(2019)(2019)]
= (2)(3)(4)...(2020)*(2020)(2021)(2022)...(4038)/[(1)(2)(3)...(2019)*(2019)(2020)(2021)...(4037)]
= 2020*2020*4038, lots of cancelling

check my thinking