Chứng tỏ $\frac{1}{3}$ + $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ + ….. + $\frac{2019}{3^{2019}}$ < $0,75$ 03/09/2021 Bởi Serenity Chứng tỏ $\frac{1}{3}$ + $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ + ….. + $\frac{2019}{3^{2019}}$ < $0,75$
@hieu Đặt : A = 1/3 +1/3² + …. + 2019/ $3^{2018}$ => 3A = 1 + 2/3 +……+ 2019/$3^{2018}$ =>2A = 1 + 1/3 +……+ 1/$3^{2018}$ – 2019/$3^{2019}$ =>6A = 3+1 + 1/3 +1/3² +……+ 1/$3^{2019}$ – 2019/$3^{2018}$ => 4A = 3- 2020 / $3^{2018}$ + 2019/$3^{2019}$ =>A = 3/4 – 4041/$3^{2019}$ ×4 < 3/4 = 0,75 Xin hay nhất ạ !! Bình luận
Đặt :` B = 1/3 + 2/3^2 + 3/3^3 + …. + 2019/3^2019` `⇒ 3B = 1 + 2/3 + …. + 2019/3^2018` `⇒ 3B – B = ( 1 + 2/3 + … + 2019/3^2018 ) – ( 1/3 + 2/3^2 + …. + 2019/3^2019 )` `⇒ 2B = 1 + 1/3 + …. + 1/3^2018 – 2019/3^2019` `⇒ 6B = 3 + 1 + 1/3 + …. + 1/3^2019 – 2019/3^2018` `⇒ 6B – 2B = ( 3 + 1 + 1/3 + … + 1/3^2019 – 2019/3^2018 ) – ( 1 + 1/3 + … + 1/3^2018 – 2019/3^2019 )` `⇒ 4B = 3 – 2020/3^2018 + 2019/3^2019` `⇒ B = 3/4 – 4041/3^2019 < 3/4` `⇒ B < 3/4` ( Điều phải chứng minh ) Bình luận
@hieu
Đặt : A = 1/3 +1/3² + …. + 2019/ $3^{2018}$
=> 3A = 1 + 2/3 +……+ 2019/$3^{2018}$
=>2A = 1 + 1/3 +……+ 1/$3^{2018}$ – 2019/$3^{2019}$
=>6A = 3+1 + 1/3 +1/3² +……+ 1/$3^{2019}$ – 2019/$3^{2018}$
=> 4A = 3- 2020 / $3^{2018}$ + 2019/$3^{2019}$
=>A = 3/4 – 4041/$3^{2019}$ ×4 < 3/4 = 0,75
Xin hay nhất ạ !!
Đặt :` B = 1/3 + 2/3^2 + 3/3^3 + …. + 2019/3^2019`
`⇒ 3B = 1 + 2/3 + …. + 2019/3^2018`
`⇒ 3B – B = ( 1 + 2/3 + … + 2019/3^2018 ) – ( 1/3 + 2/3^2 + …. + 2019/3^2019 )`
`⇒ 2B = 1 + 1/3 + …. + 1/3^2018 – 2019/3^2019`
`⇒ 6B = 3 + 1 + 1/3 + …. + 1/3^2019 – 2019/3^2018`
`⇒ 6B – 2B = ( 3 + 1 + 1/3 + … + 1/3^2019 – 2019/3^2018 ) – ( 1 + 1/3 + … + 1/3^2018 – 2019/3^2019 )`
`⇒ 4B = 3 – 2020/3^2018 + 2019/3^2019`
`⇒ B = 3/4 – 4041/3^2019 < 3/4`
`⇒ B < 3/4` ( Điều phải chứng minh )