`S = 1 . 2 + 2.3 + ... + n(n+1)`
`3S = 1.2.3 + 2.3. (4-1) + ... + n(n+1)[(n+2)-(n-1)]`
`3S = 1.2.3 + 2.3.4 - 1.2.3 + ... + n(n+1)(n+2) - (n-1)n(n+1) `
`3S = n(n+1)(n+2)`
`S = (n(n+1)(n+2))/3`
`=> n(n+1)(n+2) = 3542 . 3`
`=> n(n+1)(n+2) = 10626`
Mà `21 . 22 . 23 = 10626 `
`=> n = 21` (Thỏa mãn)
Vậy ...