Question

Chứng minh rằng :  \(\dfrac{7}{2}\) < \(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\) + .... + \(\dfrac{1}{99.100}\) < \(\dfrac{5}{6}\)

 

Answer 1

Sửa đề: Chứng minh rằng: `7/2 < 1/(1.2) + 1/(3.4) + ... + 1/(99.100) < 5/6`

Lý do sửa: `7/2 > 5/6`

Bài làm: 

Ta có: `1/(1.2) + 1/(3.4) + ... + 1/(99.100)`

`= (1/(1.2) + 1/(3.4)) + ... + 1/(99.100)`

`= (1/2 + 1/12) + ... + 1/(99.100)`

`= 7/12 + ... + 1/(99.100) > 7/12 (1) `

Lại có: `1/(1.2) + 1/(3.4) + ... + 1/(99.100)`

`= 1 - 1/2 + 1/3 - 1/4 + ... + 1/99 - 1/100`

`= (1 - 1/2 + 1/3) - 1/4 + ... + 1/99 - 1/100`

`= (6/6 - 3/6 + 2/6) - 1/4 + ... + 1/99 - 1/100`

`= 5/6 - 1/4 + ... + 1/99 - 1/100`

`= 5/6 - (1/4 - 1/5) - ... - (1/98 - 1/99) - 1/100`

Do: 

`4 < 5 => 1/4 > 1/5 => 1/4 - 1/5 > 0`

....

`98 < 99 => 1/98 > 1/99 => 1/98 - 1/99 > 0`

`=> 5/6 - (1/4 - 1/5) - ... - (1/98 - 1/99) - 1/100 < 5/6 (2) `

Từ `(1)(2) => 7/2 < 1/(1.2) + 1/(3.4) + ... + 1/(99.100) < 5/6 (đpcm)`