Toán Chứng minh rằng : A= 1/5+1/5^2+1/5^3+…..+1/5^2008 nhỏ hơn 1/4 10/07/2021 By Harper Chứng minh rằng : A= 1/5+1/5^2+1/5^3+…..+1/5^2008 nhỏ hơn 1/4
`A= 1/5+1/5^2+1/5^3+…..+1/5^(2008)` `⇔5A=1+1/5+1/5^2+1/5^3+…..+1/5^(2007)` `⇔5A-A=1+1/5+1/5^2+1/5^3+…..+1/5^(2007)-(1/5+1/5^2+1/5^3+…..+1/5^(2008))` `⇔4A=1-1/5^(2008)` `⇔A=1/4 -1/(4×5^(2008))<1/4`(ĐPCM) Trả lời
`A = 1/5+1/5^2+1/5^3+…..+1/5^2008` `=> 5.A = 5(1/5 + 1/5^2 +1/5^3+…..+1/5^2008 )` `=> 5.A = 1+ 1/5 + … + 1/5^2007` `=> 5.A – A =( 1+ 1/5 + … + 1/5^2007) – ( 1/5 + … + 1/5^2008 )` `=> 4A = 1 – 1/5^2008` `=> A = 1 – 1/5^2008 : 4` `=> A = 1/4 – 1/(5^2008 . 4) < 1/4` `=> A < 1/4` Trả lời
`A= 1/5+1/5^2+1/5^3+…..+1/5^(2008)`
`⇔5A=1+1/5+1/5^2+1/5^3+…..+1/5^(2007)`
`⇔5A-A=1+1/5+1/5^2+1/5^3+…..+1/5^(2007)-(1/5+1/5^2+1/5^3+…..+1/5^(2008))`
`⇔4A=1-1/5^(2008)`
`⇔A=1/4 -1/(4×5^(2008))<1/4`(ĐPCM)
`A = 1/5+1/5^2+1/5^3+…..+1/5^2008`
`=> 5.A = 5(1/5 + 1/5^2 +1/5^3+…..+1/5^2008 )`
`=> 5.A = 1+ 1/5 + … + 1/5^2007`
`=> 5.A – A =( 1+ 1/5 + … + 1/5^2007) – ( 1/5 + … + 1/5^2008 )`
`=> 4A = 1 – 1/5^2008`
`=> A = 1 – 1/5^2008 : 4`
`=> A = 1/4 – 1/(5^2008 . 4) < 1/4`
`=> A < 1/4`