Tính phép Tính sau
(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]
Tính phép Tính sau (1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]
By Parker
By Parker
Tính phép Tính sau
(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]
Đặt:
`A = 1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(99.100)`
`A = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/99 – 1/100`
`A = 1 – 1/100`
`A = 99/100`
Đặt:
`B = (1 – 1/2) . (1 – 1/3) . (1 – 1/4) . … . (1 – 1/2022)`
`B = 1/2 . 2/3 . 3/4 . … . 2021/2022`
`B = (1 . 2 . 3 . … . 2021)/(2 . 3 . 4 . … . 2022)`
`B = 1/2022`
Như vậy:
`(1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(99.100)) . [(1 – 1/2) . (1 – 1/3) . (1 – 1/4) . … . (1 – 1/2022)]
`= 99/100 . 1/2022`
`= 99/202200`
Cách 1:Chi tiết,dễ hiểu
`(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]`
Đặt :
`A=1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100`
`B=(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)`
`=>(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]=A.B`
Theo bài ra ta có:
`A=1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100`
Áp dụng công thức `1/[n.(n+1)]=1/n-1/[n+1]`
`=>A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100`
`A=1-1/100`
`A=100/100-1/100`
`A=99/100`
Vậy `A=99/100(1)`
Lại có:
`B=(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)`
`B=(2/2-1/2)(3/3-1/3)(4/4-1/4)…(2022/2022-1/2022)`
`B=1/2. 2/3. 3/4…. 2021/2022`
`B=1/2022`
Vậy `B=1/2022(2)`
Từ `(1),(2)`
`=>A.B=99/100. 1/2022`
`=>A.B=99/202200`
`=>(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]=99/202200`
Cách 2:Ngắn gọn,hơi khó hiểu với một số bạn.
`(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]`
`=(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100).[(2/2-1/2)(3/3-1/3)(4/4-1/4)…(2022/2022-1/2022)]`
`=(1-1/100).[1/2. 2/3. 3/4…. 2021/2022]`
`=(100/100-1/100). 1/2022`
`=99/100. 1/2022`
`=99/202200`