CMR : `C =1/5 + 1/5^2 + 1/5^3 + … + 1/5^2017 < 1/4` 21/08/2021 Bởi Lydia CMR : `C =1/5 + 1/5^2 + 1/5^3 + … + 1/5^2017 < 1/4`
Đáp án: `C = 1/5 + 1/5^2 + 1/5^3 + … + 1/5^{2017}` `-> 1/5C = 1/5^2 + 1/5^3 + … + 1/5^{2018}` `->C – 1/5C = (1/5 + 1/5^2 + 1/5^3 + … + 1/5^{2017}) – (1/5^2 + 1/5^3 + … + 1/5^{2018})` `-> 4/5C = 1/5 – 1/5^{2018}` `-> C = 1/4 – 1/(5^{2017} . 4)` `text{Ta thấy :}` `1/4 – 1/(5^{2017} . 4) < 1/4` `-> C < 1/4 (đpcm)` Bình luận
`C=1/5 + 1/5^2 + 1/5^3 +…+1/5^2017` `1/5 C= 1/5 ( 1/5 +1/5^2 + 1/5^3+…+1/5^2017)` `1/5C = 1/5^2 + 1/5^3 + 1/5^4 +…+ 1/5^2018` `C- 1/5 C= 1/5 + 1/5^2 + 1/5^3 +…+1/5^2017 – 1/5^2 – 1/5^3 – 1/5^4-..-1/5^2018` `4/5 C = 1/5 – 1/5^2018` `C= (1/5 – 1/5^2018):4/5` `C= (1/5- 1/5^2018) . 5/4` `C = 1/4 – 1/(5^2017 .4) < 1/4` Vậy `C < 1/4` Bình luận
Đáp án:
`C = 1/5 + 1/5^2 + 1/5^3 + … + 1/5^{2017}`
`-> 1/5C = 1/5^2 + 1/5^3 + … + 1/5^{2018}`
`->C – 1/5C = (1/5 + 1/5^2 + 1/5^3 + … + 1/5^{2017}) – (1/5^2 + 1/5^3 + … + 1/5^{2018})`
`-> 4/5C = 1/5 – 1/5^{2018}`
`-> C = 1/4 – 1/(5^{2017} . 4)`
`text{Ta thấy :}` `1/4 – 1/(5^{2017} . 4) < 1/4`
`-> C < 1/4 (đpcm)`
`C=1/5 + 1/5^2 + 1/5^3 +…+1/5^2017`
`1/5 C= 1/5 ( 1/5 +1/5^2 + 1/5^3+…+1/5^2017)`
`1/5C = 1/5^2 + 1/5^3 + 1/5^4 +…+ 1/5^2018`
`C- 1/5 C= 1/5 + 1/5^2 + 1/5^3 +…+1/5^2017 – 1/5^2 – 1/5^3 – 1/5^4-..-1/5^2018`
`4/5 C = 1/5 – 1/5^2018`
`C= (1/5 – 1/5^2018):4/5`
`C= (1/5- 1/5^2018) . 5/4`
`C = 1/4 – 1/(5^2017 .4) < 1/4`
Vậy `C < 1/4`