Tính tổng: 1^2 + 3^2 + 5^2 + + (2n - 1)^2
Tính tổng: 12 + 32 + 52 + … + (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 = \(\frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}\).