Tính F= (1-1/1+2)(1-1/1+2+3)........(1-1/1+2+3+...+n)

`F=(1-1/(1+2))(1-1/(1+2+3))...(1-1/(1+2+3+...+n))`

`=(1-1/((1+2).2:2))(1-1/((1+3).3:2))...(1-1/((1+n).n:2))`

`=(1-2/(2.3))(1-2/(3.4))(1-2/(4.5))...(1-2/(n(n+1)))`

`=(2.3-2)/(2.3).(3.4-2)/(3.4).(4.5-2)/(4.5) ... (n(n+1)-2)/(n(n+1))`

Ta có: `n(n+1)-2=n^2+n-2=(n^2-n)+(2n-2)=n(n-1)+2(n-1)=(n+2)(n-1)`

`=>` `F=((2+2)(2-1))/(2.3).((3+2)(3-1))/(3.4).((4+2)(4-1))/(4.5) ... ((n+2)(n-1))/(n(n+1))`

`=(1.4)/(2.3).(2.5)/(3.4).(3.6)/(4.5) ... ((n-1)(n+2))/(n(n+1))`

`=((1.2.3...(n-1)).(4.5.6...(n+2)))/((2.3.4...n).(3.4.5...(n+1)))`

`=1/n . (n+2)/3`

`=(n+2)/(3n)`