Tính tổng: S = 1^2 + 2^2 + 3^2 + + n^2

Ta có:

n² – n = n(n – 1)

⇔ n² = (n – 1)n + n

Khi đó:

S = 1² + 2² + 3² + ... + n²

S = 1 + 1 . 2 + 2 . 3 + ... + (n – 1)n + n

S = [1 . 2 + 2 . 3 + ... + (n – 1)n] + (1 + 2 + ... + n)

\[{\rm{S}} = \frac{{\left( {n - 1} \right)n\left( {n + 1} \right)}}{3} + \frac{{n\left( {n + 1} \right)}}{2}\]

\[{\rm{S}} = \frac{{2\left( {n - 1} \right)n\left( {n + 1} \right) + 3n\left( {n + 1} \right)}}{6}\]

\[{\rm{S}} = \frac{{n\left( {n + 1} \right)\left( {2n - 2 + 3} \right)}}{6}\]

\[{\rm{S}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\].