Đặt `S = 1^3 + 2^3 + 3^3 + ... + n^3`
`= 1^2 . 1 + 2^2 . 2 + 3^2 . 3 + ... + n^2 . n`
`= 1 . (2 - 1) + 2^2 . (3 - 1) + 3^2 . (4 - 1) + ... + n^2 . (n + 1 - 1) `
`= 1 . 2 - 1 + 2^2 . 3 - 2^2 + 3^2 . 4 - 3^2 + ... + n^2 . (n+1) - n^2`
`= [1. 2 + 2^2 . 3 + 3^2 . 4 + ... + n^2 (n+1)] - (1 + 2^2 + 3^2 + ... + n^2)`
Đặt `A = 1. 2 + 2^2 . 3 + 3^2 . 4 + ... + n^2 (n+1)`
`A = 1.2 + (1+1).2 . 3 + (2 + 1) . 3 . 4 + ... + (n - 1 + 1) . n . (n+1) `
`= 1 . 2 + 1.2.3 + 2.3 + 2.3.4 + 3.4 + ... + (n-1) . n . (n+1) + n(n+1) `
`= [1 . 2 + 2.3 + 3.4 + ... + n . (n+1)] + [(1.2.3 + 2.3.4 + ... + (n-1) . n . (n+1)]`
Đặt `A_1 = 1 . 2 + 2.3 + 3.4 + ... + n . (n+1)`
`3A_1 = 1.2.3 + 2.3. (4-1) + 3.4.(5-2) + ... + n(n+1).[(n+2) - (n-1)]`
`3A_1 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n(n+1)(n+2) - (n-1)n(n+1)`
`3A_1 = n(n+1)(n+2)`
`A_1 = (n(n+1)(n+2))/3`
Đặt `A_2 = 1.2.3 + 2.3.4 + ... + (n-1) . n . (n+1)`
`4A_2 = 1.2.3.4 + 2.3.4.(5-1) + ... + (n-1)n(n+1)[(n+2)-(n-2)]`
`4A_2 = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + (n-1).n.(n+1)(n+2) - (n-2)(n-1)n.(n+1)`
`4A_2 = (n-1).n.(n+1)(n+2)`
`A_2 = ( (n-1).n.(n+1)(n+2))/4`
Khi đó `A = A_1 + A_2 = (n(n+1)(n+2))/3 + ((n-1).n.(n+1)(n+2))/4`
Đặt `B = 1 + 2^2 + 3^2 + ... + n^2`
`= 1 . (2-1) + 2. (3-1) + 3.(4-1) + ... + n.(n+1 - 1) `
`= [1 . 2 + 2.3 + 3.4 +... + n(n+1)] - (1+2+3+...+n)`
`= (n(n+1)(n+2))/3 - (n(n+1))/2`
`= n(n+1). ((n+2)/3 - 1/2)`
`= (n(n+1)(2n+1))/6`
Khi đó `S = A - B`
`= (n(n+1)(n+2))/3 + ((n-1).n.(n+1)(n+2))/4 - (n(n+1)(2n+1))/6`
`= n(n+1) [(n+2)/3 + ((n-1)(n+2))/4 - (2n + 1)/6]`
`= n(n+1) [(n+2)/3 + (n^2 + n - 2)/4 - (2n + 1)/6]`
`= n(n+1) [(4n+8)/12 + (3n^2 + 3n - 6)/12 - (4n + 2)/12]`
`= n(n+1) (4n+8 +3n^2 + 3n - 6 - 4n - 2)/12`
`= n(n+1) (3n^2 + 3n)/12`
`= n(n+1) (n^2 + n)/4`
`= n(n+1) (n(n+1))/4`
`= ((n.(n+1))/2)^2`
Mà `1 + 2 + 3+ ....+ n = (n.(n+1))/2 `
Nên `S = (1 + 2 + 3+ ....+ n)^2 (đpcm)`