Tính tổng: 1^2 + 3^2 + 5^2 + + (2n - 1)^2

Đặt S(2n – 1) = 1² + 3² + 5² + … + (2n – 1)²

S(2n) = 2² + 4² + 6² + … + (2n)² = \(\frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)

Suy ra: S(2n – 1) + S(2n) = 1² + 2² + 3² + 4² + 5² + … + (2n)² + (2n – 1)²

Ta có: 1² + 2² + 3² + 4² + 5² + … n² = \(\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)

S(2n – 1) + S(2n) = \(\frac{{2n\left( {2n + 1} \right)\left( {4n + 1} \right)}}{6}\)

S(2n – 1) = \(\frac{{2n\left( {2n + 1} \right)\left( {4n + 1} \right)}}{6} - \frac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}\)

Vậy 1² + 3² + 5² + … + (2n – 1)² = \(\frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}\).