Chứng minh rằng A= 1.2.3.4….2019.2020.(1+$\frac{1}{2}$ +$\frac{1}{3}$+…+ $\frac{1}{2019}$ +$\frac{1}{2020}$) chia hết cho 2021

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1. Giải thích các bước giải:

   Ta có:

   $A=1.2.3…2019.2020.(1+\dfrac12+\dfrac13+…+\dfrac1{2019}+\dfrac1{2020})$ 

   $\to A=1.2.3…2019.2020.((1+\dfrac1{2020})+(\dfrac12+\dfrac1{2019})+…+(\dfrac1{1010}+\dfrac1{1011}))$ 

   $\to A=1.2.3…2019.2020.(\dfrac{2020+1}{2020}+\dfrac{2+2019}{2.2019}+…+\dfrac{1010+1011}{1010.1011})$ 

   $\to A=1.2.3…2019.2020.(\dfrac{2021}{2020}+\dfrac{2021}{2.2019}+…+\dfrac{2021}{1010.1011})$ 

   $\to A=2021.1.2.3…2019.2020.(\dfrac{1}{2020}+\dfrac{1}{2.2019}+…+\dfrac{1}{1010.1011})$ 

   $\to A\quad\vdots\quad 2021$

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