Giải thích các bước giải:
Ta có `A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100`
`A = (1/1.2 + 1/3.4) + (1/5.6 + ... + 1/99.100)`
`A = 7/12 + (1/5.6 + ... + 1/99.100)>7/12` `(`vì `1/5.6 + ... + 1/99.100>0``)`
Ta có `A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100`
`=>A = 1/1-1/2 + 1/3-1/4 + 1/5-1/6 + ... + 1/99-1/100`
`=>A = (1/1 + 1/3 + 1/5 + ... + 1/99)-(1/2+1/4+1/6+...+1/100)`
`=>A = (1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100)-2(1/2+1/4+1/6+...+1/100)`
`=>A = (1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100)-(1+1/2+1/3+...+1/500)`
`=>A=1/51+1/52+1/53+...+1/100`
Tổng `A` có:
`(100-51):1+1=50` (số hạng)
Như vậy, ta nhóm `10` số vào `1` nhóm được:
`A=(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)+(1/71+1/72+...+1/80)+(1/81+1/82+...+1/90)+(1/91+1/92+...+1/100)`
Ta thấy:
`(1/51+1/52+...+1/60)<10. 1/50=1/5`
`(1/61+1/62+...+1/70)<10. 1/60=1/6`
`(1/71+1/72+...+1/80)<10. 1/70=1/7`
`(1/81+1/82+...+1/90)<10. 1/80=1/8`
`(1/91+1/92+...+1/100)<1/9`
`=>(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)+(1/71+1/72+...+1/80)+(1/81+1/82+...+1/90)+(1/91+1/92+...+1/100)<1/5+1/6+1/7+1/8+1/9<5/6`
`=>A<5/6`
Vậy `7/12<A<5/6`
