Đặt `A = 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + 1/(1 + 2 + 3 + 4) + ... + 1/(1 + 2 + 3 + 4 + ... + 2021)`
Ta có:
`1/(1 + 2) = 1/3 = 2/(2. 3)`
`1/(1 + 2 + 3) = 1/6 = 2/12 = 2/(3. 4)`
`1/(1 + 2 +3 + 4) = 1/10 = 2/20 = 2/(4. 5)`
`...`
`1/(1 + 2 + 3 + 4 + ... + 2021) = 1/2043231 = 2/4086462 = 2/(2021. 2022)`
`=> A = 1 + 2/(2. 3) + 2/(3. 4) + 2/(4. 5) + ... + 2/(2021. 2022)`
`=> A = 1 + 2. (1/(2. 3) + 1/(3. 4) + 1/(4. 5) + ... + 1/(2021. 2022))`
`=> A = 1 + 2. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2021 - 1/2022)`
`=> A + 1 + 2. (1/2 - 1/2022)`
`=> A = 1 + 1 - 1/1011`
`=> A = 2 - 1/1011`
`=> A = 2022/1011 - 1/1011`
`=> A = 2021/1011`
Vậy `A = 2021/1011`
