Giải thích các bước giải:
Ta có:
$A=1.2.3...2019.2020.(1+\dfrac12+\dfrac13+...+\dfrac1{2019}+\dfrac1{2020})$
$\to A=1.2.3...2019.2020.((1+\dfrac1{2020})+(\dfrac12+\dfrac1{2019})+...+(\dfrac1{1010}+\dfrac1{1011}))$
$\to A=1.2.3...2019.2020.(\dfrac{2020+1}{2020}+\dfrac{2+2019}{2.2019}+...+\dfrac{1010+1011}{1010.1011})$
$\to A=1.2.3...2019.2020.(\dfrac{2021}{2020}+\dfrac{2021}{2.2019}+...+\dfrac{2021}{1010.1011})$
$\to A=2021.1.2.3...2019.2020.(\dfrac{1}{2020}+\dfrac{1}{2.2019}+...+\dfrac{1}{1010.1011})$
$\to A\quad\vdots\quad 2021$
