D=1/2018(2/1 +3/2 + 4/3 + 5/4 +…+2019/2018)

0 bình luận về “D=1/2018(2/1 +3/2 + 4/3 + 5/4 +…+2019/2018)”

1. D = $\dfrac{1}{2018}$ ( $\dfrac{2}{1}$ + $\dfrac{3}{2}$ + $\dfrac{4}{5}$ + …. + $\dfrac{2019}{2018}$

   = $\dfrac{1}{2018}$ ( 2018 + 1 + $\dfrac{1}{2}$ + $\dfrac{1}{5}$ + …. + $\dfrac{1}{2018}$

   `= 1 +` $\dfrac{1}{2018}$ `( 1 +` $\dfrac{1}{2}$ `+` $\dfrac{1}{5}$ `+ …. +` $\dfrac{1}{2018}$

   Tách `1 +` $\dfrac{1}{2}$ `+` $\dfrac{1}{5}$ `+ …. +` $\dfrac{1}{2018}$ = C

   `⇒ C = 1 +` ($\dfrac{1}{2}$ `+` $\dfrac{1}{5}$ `+ …. +` $\dfrac{1}{2018}$)

   Ta có: $\dfrac{1}{2}$ `>` $\dfrac{1}{2019}$

   $\dfrac{1}{3}$ `>` $\dfrac{1}{2019}$

   `……`

   $\dfrac{1}{2018}$ `>` $\dfrac{1}{2019}$

   `⇒ 1` `+` ($\dfrac{1}{2}$ + $\dfrac{1}{5}$ `+ …. +` $\dfrac{1}{2018}$) `( 2017` số`)`

   `= 1 +` $\dfrac{1}{2019}$.2017

   `= 1 +` $\dfrac{2017}{2019}$

   Khi đó:

   `D= 1 `+ $\dfrac{1}{2018}$`.(1 +` $\dfrac{2017}{2019}$`)`

   `= 1 +` $\dfrac{1}{2018}$ `+` $\dfrac{2017}{2019}$

   `=` $\dfrac{2019}{2018}$ `+` $\dfrac{2017}{2019}$

   Bình luận

2. D=1/2018(2/1 +3/2 + 4/3 + 5/4 +…+2019/2018)

   D=1/2018((1+1/1)+(1+1/2)+….+(1+1/2018))

   D=1/2018((1+1+….+1)+(1/1+1/2+1/3+…1/2018))

   D=1/2018(1*2018+(1/1+1/2+1/3+…1/2018))

   D=1/2018(2018+(1/1+1/2+1/3+…1/2018))

   D=1+1/2018(1/1+1/2+1/3+…1/2018)

   D-1=1/2018(1/1+1/2+1/3+…1/2018)

   (D-1)*2018=1/1+1/2+1/3+…1/2018

   Bình luận