Cho A = 1/(1.2) + 1/(3.4) + 1/(5.6) + ... + 1/(99.100) Chứng minh rằng: 7/12 < A < 5/6

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`