`(2.2020)/(1+1/(1+2) + 1/(1+2+3) + .............+ 1/(1+2+2+............+2020))`
Ta có : 1 + `1/(1+2)` + `1/(1+2+3)`+..............+ `1/(1+2+3+.........+2020)`
= `2/(1.2)` + `2/(2(1+2))` + `2/(2(1+2+3))`+..............+ `2/(2(1+2+3+.........+2020))`
= `2/2` + `2/6` + `2/12` + ............... + `2/4082420`
= `2/(1.2)` + `2/(2.3)` + `2/(3.4)` + ................. + `2/(2020.2021)`
Đặt A = `2/(1.2)` + `2/(2.3)` + `2/(3.4)` + ................. + `2/(2020.2021)`
`1/2` A = `1/(1.2)` + `1/(2.3)` + `1/(3.4)` + ................. + `1/(2020.2021)`
`1/2` A = `(2-1)/(1.2)` + `(3-2)/(2.3)` + `(4-3)/(3.4)` + ................. + `(2021-2020)/(2020.2021)`
`1/2` A = 1 - `1/2` + `1/2` - `1/3` + ................ + `1/2020` - `1/2021`
`1/2` A = 1 - `1/2021` = `2020/2021`
⇒ A = `2020/2021` : `1/2`
⇒ A = `4040/2021`
Vậy `(2.2020)/(1+1/(1+2) + 1/(1+2+3) + .............+ 1/(1+2+2+............+2020))`
= 4040 : `4040/2021`
= 2021
