`b)`
Ta có: `n^3 - n = n(n^2 - 1) = n(n^2 - 1^2) = n(n - 1)(n + 1)`
`=> n^3 - n = (n - 1)n(n + 1)`
`=> n^3 = (n - 1)n(n + 1) + n`
Áp dụng công thức trên, ta có:
`1^3 = 1`
`2^3 = (2 - 1). 2. (2 + 1) + 2 = 1. 2. 3 + 2`
`3^3 = (3 - 1). 3. (3 + 1) + 3 = 2. 3. 4 + 3`
`...`
`n^3 = (n - 1)n(n + 1) + n`
`=> 1^3 + 2^3 + 3^3 + ... + n^3`
`= 1 + 1. 2. 3 + 2 + 2. 3. 4 + 3 + ... + (n - 1)n(n + 1) + n`
`=> S = (1 + 2 + 3 + ... + n) + [1. 2. 3 + 2. 3. 4 + ... + (n - 1)n(n + 1)]`
$\text{Đặt A = 1 + 2 + 3 + ... + n; B = 1. 2. 3 + 2. 3. 4 + ... + (n - 1)n(n + 1)}$
`+) => A = (n + 1). [(n - 1)/1 + 1] : 2`
`= (n + 1). n : 2`
`+) => 4B = 1. 2. 3. 4 + 2. 3. 4. 4 + ... + (n - 1)n(n + 1). 4`
$\text{= 1. 2. 3. 4 + 2. 3. 4(5 - 1) + ... + (n - 1)n(n + 1)[(n + 2) - (n - 2)]}$
$\text{= 1. 2. 3. 4 + 2. 3. 4. 5 - 1. 2. 3. 4 + ... + (n - 1)n(n + 1)(n + 2) - (n - 1)n(n + 1)(n - 2)}$
`= (n - 1)n(n + 1)(n + 2)`
`=> B = ((n - 1)n(n + 1)(n + 2))/4`
`=> S = (n + 1). n : 2 + ( (n - 1)n(n + 1)(n + 2) )/4`
`= (2(n^2 + n))/4 + ( (n^2 - 1)(n^2 + 2n) )/4`
`= (2(n^2 + n) + (n^2 - 1)(n^2 + 2n))/4`
`= (2n^2 + 2n + n^4 + 2n^3 - n^2 - 2n)/4`
`= (n^4 + 2n^3 + n^2)/4`
Vậy `S = (n^4 + 2n^3 + n^2)/4`
