có công thức: 1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6 (*)
đặt: S(2n + 1) = 1^2 + 3^2 + 5^2 + ... + (2n+1)^2
S(2n) = 2^2 + 4^2 + ...+ (2n)^2 = 2^2(1^2 + 2^2 + 3^2 +...+ n^2) = 4n(n+1)(2n+1)/6
=> S(2n+1) + S(2n) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + ... + (2n)^2 + (2n+1)^2
áp dụng (*) ta có: S(2n+1) + S(2n) = (2n+1)[(2n+1) + 1)[2(2n+1) + 1)]/6 = (2n+1)(2n+2)(4n+3)/6
=> S(2n+1) = (2n+1)(2n+2)(4n+3)/6 - S(2n) = 2(2n+1)(n+1)(4n+3)/6 - 4n(n+1)(2n+1)/6
= (2n+1)(n+1)/3.[(4n +3) - 2n] = (2n+1)(n+1)(2n+3)/3 vậy S(2n+1) = (2n+1)(n+1)(2n+3)/3