Q = 1.2.3.4….2020 . ( 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + …+ $\frac{1}{2020}$ ) Chứng tỏ Q chia hết cho 2021

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1. Đáp án + giải thích bước giải :

   Ta có :

   `1 + 1/2 + 1/3 + ….. + 1/2020`

   `= (1 + 1/2020) + (1/2 + 2019) + (1/3 + 2018) + …. + (1/1010 + 1/1011)`

   `= 2021/2020 + 2021/4038 + … + 2021/1021110`

   `= 2021/(1 . 2020) + 2021/(2 . 2019) + …. + 2021/(1010 . 1011)`

   `= 2021 . (1/(1 . 2020) + 1/(2 . 2019) + …. + 1/(1010 . 1011) )`

   Theo đề : `Q = 1 . 2 . 3 …. 2020 . (1 + 1/2 + 1/3 + …. + 1/2020)`

   `⇔ Q = 1 . 2 . 3 ….. 2020 . 2021 . (1/(1 . 2020) + 1/(2 . 2019) + … + 1/(1010 . 1011) )`

   `⇔ 1 . 2 . 3 ….. 2020 . 2021 . (1/(1 . 2020) + 1/(2 . 2019) + … + 1/(1010 . 1011) ) \vdots 2021`

   `⇔ Q = 1 . 2 . 3 … 2020 . (1 + 1/2 + 1/3 + … .+ 1/2020) \vdots 2021`

    

   Trả lời

2. Giải thích các bước giải:

   Ta có:

   $1+\dfrac12+\dfrac13+…+\dfrac1{2020}$

   $=(1+\dfrac1{2020})+(\dfrac12+\dfrac1{2019})+…+(\dfrac{1}{1010}+\dfrac1{1011})$

   $=\dfrac{2021}{1.2020}+\dfrac{2021}{2.2019}+…+\dfrac{2021}{1010.1011}$

   $=2021(\dfrac{1}{1.2020}+\dfrac{1}{2.2019}+…+\dfrac{1}{1010.1011})$

   Mà $1.2.3…2020.(\dfrac{1}{1.2020}+\dfrac{1}{2.2019}+…+\dfrac{1}{1010.1011})\in Z$

   $\to 2021(\dfrac{1}{1.2020}+\dfrac{1}{2.2019}+…+\dfrac{1}{1010.1011}).1.2.3…2020\quad\vdots\quad 2021$

   $\to Q\quad\vdots\quad 2021$

   Trả lời