Tính tổng 12 + 22 + … + n2.

Tính tổng 1² + 2² + … + n².

F = 1² + 2² + 3² + … + n²

F = 1 + (1 + 1).2 + (1 + 2).3 + (1 + 3).4 + … + (1 + n – 1)n

F = 1 + (2 + 1.2) + (3 + 2.3) + (4+ 3.4) + … + [n + (n – 1)n]

F = (1 + 2 + 3 + 4 + … + n) + [1.2 + 2.3 + 3.4 + …. + (n – 1)n]

Đặt A = 1 + 2 + 3 + 4 + … + n thì A = $\frac{n\left(n+1\right)}{2} \left(1\right)$

Đặt B = [1.2 + 2.3 + 3.4 + …. + (n – 1)n]

Xét 3B = 1.2.3 + 2.3.3 + 3.4.3 + … + (n – 1).n.3

3B = [1.2.3 + 2.3.4 + … + (n – 1).n.(n + 1)] – (1.2.3 + 2.3.4 + … + (n – 2)(n – 1)n)

3B = (n – 1)n(n + 1)

B = $\frac{\left(n–1\right)n\left(n+1\right)}{3}\left(2\right)$

Từ (1) và (2) suy ra:

F = $\frac{n\left(n+1\right)}{2}+\frac{\left(n–1\right)n\left(n+1\right)}{3}=\frac{3n^2+3n+2n\left(n^2-1\right)}{6}=\frac{2n^3+3n^2+n}{6}$

$=\frac{n\left(2n^2+3n+1\right)}{6}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Vậy F = 1² + 2² + 3² + … + n² $=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$