chứng tỏ rằng: 3/1.2.3+5/2.3.4+7/3.4.5+…+ 2017/1008.1009.1010<5/4 30/07/2021 Bởi Eden chứng tỏ rằng: 3/1.2.3+5/2.3.4+7/3.4.5+…+ 2017/1008.1009.1010<5/4
Gọi `A = 3/(1.2.3) + 5/(2.3.4) + 7/(3.4.5) + … + 2017/(1008.1009.1010)` `⇒ A = 2/(1.2.3) + 1/(1.2.3) + 4/(2.3.4) + 1/(2.3.4) + 6/(3.4.5) + 1/(3.4.5) + … + 2016/(1008.1009.1010) + 1/(1008.1009.1010)` `= ( 2/(1.2.3) + 4/(2.3.4) + 6/(3.4.5) + … + 2016/(1008.1009.1010) ) + ( 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + … + 1/(1008.1009.1010) )` `= ( (1.2)/(1.2.3) + (2.2)/(2.3.4) + (3.2)/(3.4.5) + … + (1008.2)/(1008.1009.1010) ) + 1/2.(2/(1.2.3) + 2/(2.3.4) + 2/(3.4.5) + … + 2/(1008.1009.1010) )` `= ( 2/(2.3) + 2/(3.4) + 2/(4.5) + … + 2/(1009.1010) ) + 1/2. ( 1/(1.2) – 1/(2.3) + 1/(2.3) – 1/(3.4) + 1/(3.4) – 1/(4.5) + … + 1/(1008.1009) – 1/(1009.1010) )` `= 2. ( 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(1009.1010) ) + 1/2. ( 1/(1.2) – 1/(1009.1010) )` `= 2. ( 1/2 – 1/3 + 1/3 -1/4 + 1/4 – 1/5 + … + 1/1009 – 1/1010) + 1/2. ( 1 – 1/2 – 1/(1009.1010)) ` `= 2. (1/2 – 1/1010) + 1/2. (1/2 – 1/(1009.1010) )` Vì `1/2 – 1/1010 < 1/2; 1/2 – 1/(1009.1010) < 1/2` `⇒ 2. (1/2 – 1/1010) < 1; 1/2. (1/2 – 1/(1009. 1010)) < 1/4` `⇒ 2. (1/2 – 1/1010) + 1/2. (1/2 – 1/(1009. 1010) ) < 1 + 1/4` `⇒ A < 5/4` `⇒đpcm` Bình luận
Giải thích các bước giải: Đặt `A = 3/(1.2.3) + 5/(2.3.4) + 7/(3.4.5) +…+ 2017/(1008.1009.1010)` `=> A = (2/(1.2.3) + 1/(1.2.3)) + (4/(2.3.4) + 1/(2.3.4)) + (6/(3.4.5) + 1/(3.4.5)) + … + (2016/(1008.1009.1010) + 1/(1008.1009.1010))` `=> A = (2/(1.2.3) + 4/(2.3.4) +6/(3.4.5) +….+ 2016/(1008.1009.1010))+(1/(1.2.3) + 1/(2.3.4)+ 1/(3.4.5)+…+1/(1008.1009.1010))` `=> A = ((1.2)/(1.2.3)+ (2.2)/(2.3.4) +(2.3)/(3.4.5) +….+ (1008.2)/(1008.1009.1010)) + 1/2.(2/(1.2.3) + 2/(2.3.4)+ 2/(3.4.5)+…+2/(1008.1009.1010))` `=> A = ( 2/2.3 + 2/3.4 + 2/4.5 + … + 2/1009.10010) + 1/2. (1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 + 1/3.4 – 1/4.5 + … + 1/1008.1009 – 1/1009.1010)` `=> A = 2.(1/2.3 + 1/3.4 + 1/4.5 + … + 1/1009.1010)+ 1/2. (1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 + 1/3.4 – 1/4.5 + … + 1/1008.1009 – 1/1009.1010)` `=> A = 2.(1/2-1/3+1/3-1/4+1/4-1/5+…+1/1009-1/1010) + 1/2.(1/1.2 – 1/1009.1010)` `=> A = 2.(1/2-1/1010) + 1/2.(1/2-1/1009.1010)` Vì `2.(1/2-1/1010) = 2 . 1/2 – 2 . 1/1010 < 2 . 1/2` `1/2.(1/2-1/1009.1010) = 1/2 . 1/2 – 1/2 . 1/1009.1010 < 1/2 . 1/2` `=> A < 2 . 1/2 + 1/2 . 1/2` `=> A < 1+1/4 = 5/4` `=> A < 5/4(đpcm)` Bình luận
Gọi `A = 3/(1.2.3) + 5/(2.3.4) + 7/(3.4.5) + … + 2017/(1008.1009.1010)`
`⇒ A = 2/(1.2.3) + 1/(1.2.3) + 4/(2.3.4) + 1/(2.3.4) + 6/(3.4.5) + 1/(3.4.5) + … + 2016/(1008.1009.1010) + 1/(1008.1009.1010)`
`= ( 2/(1.2.3) + 4/(2.3.4) + 6/(3.4.5) + … + 2016/(1008.1009.1010) ) + ( 1/(1.2.3) + 1/(2.3.4) + 1/(3.4.5) + … + 1/(1008.1009.1010) )`
`= ( (1.2)/(1.2.3) + (2.2)/(2.3.4) + (3.2)/(3.4.5) + … + (1008.2)/(1008.1009.1010) ) + 1/2.(2/(1.2.3) + 2/(2.3.4) + 2/(3.4.5) + … + 2/(1008.1009.1010) )`
`= ( 2/(2.3) + 2/(3.4) + 2/(4.5) + … + 2/(1009.1010) ) + 1/2. ( 1/(1.2) – 1/(2.3) + 1/(2.3) – 1/(3.4) + 1/(3.4) – 1/(4.5) + … + 1/(1008.1009) – 1/(1009.1010) )`
`= 2. ( 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(1009.1010) ) + 1/2. ( 1/(1.2) – 1/(1009.1010) )`
`= 2. ( 1/2 – 1/3 + 1/3 -1/4 + 1/4 – 1/5 + … + 1/1009 – 1/1010) + 1/2. ( 1 – 1/2 – 1/(1009.1010)) `
`= 2. (1/2 – 1/1010) + 1/2. (1/2 – 1/(1009.1010) )`
Vì `1/2 – 1/1010 < 1/2; 1/2 – 1/(1009.1010) < 1/2`
`⇒ 2. (1/2 – 1/1010) < 1; 1/2. (1/2 – 1/(1009. 1010)) < 1/4`
`⇒ 2. (1/2 – 1/1010) + 1/2. (1/2 – 1/(1009. 1010) ) < 1 + 1/4`
`⇒ A < 5/4`
`⇒đpcm`
Giải thích các bước giải:
Đặt `A = 3/(1.2.3) + 5/(2.3.4) + 7/(3.4.5) +…+ 2017/(1008.1009.1010)`
`=> A = (2/(1.2.3) + 1/(1.2.3)) + (4/(2.3.4) + 1/(2.3.4)) + (6/(3.4.5) + 1/(3.4.5)) + … + (2016/(1008.1009.1010) + 1/(1008.1009.1010))`
`=> A = (2/(1.2.3) + 4/(2.3.4) +6/(3.4.5) +….+ 2016/(1008.1009.1010))+(1/(1.2.3) + 1/(2.3.4)+ 1/(3.4.5)+…+1/(1008.1009.1010))`
`=> A = ((1.2)/(1.2.3)+ (2.2)/(2.3.4) +(2.3)/(3.4.5) +….+ (1008.2)/(1008.1009.1010)) + 1/2.(2/(1.2.3) + 2/(2.3.4)+ 2/(3.4.5)+…+2/(1008.1009.1010))`
`=> A = ( 2/2.3 + 2/3.4 + 2/4.5 + … + 2/1009.10010) + 1/2. (1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 + 1/3.4 – 1/4.5 + … + 1/1008.1009 – 1/1009.1010)`
`=> A = 2.(1/2.3 + 1/3.4 + 1/4.5 + … + 1/1009.1010)+ 1/2. (1/1.2 – 1/2.3 + 1/2.3 – 1/3.4 + 1/3.4 – 1/4.5 + … + 1/1008.1009 – 1/1009.1010)`
`=> A = 2.(1/2-1/3+1/3-1/4+1/4-1/5+…+1/1009-1/1010) + 1/2.(1/1.2 – 1/1009.1010)`
`=> A = 2.(1/2-1/1010) + 1/2.(1/2-1/1009.1010)`
Vì `2.(1/2-1/1010) = 2 . 1/2 – 2 . 1/1010 < 2 . 1/2`
`1/2.(1/2-1/1009.1010) = 1/2 . 1/2 – 1/2 . 1/1009.1010 < 1/2 . 1/2`
`=> A < 2 . 1/2 + 1/2 . 1/2`
`=> A < 1+1/4 = 5/4`
`=> A < 5/4(đpcm)`