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)`