n=1时 s=1^2=1,n(2n+1)(2n-1)/3=1*3*1/3=1,命题成立
n=2时 s=1^2 + 3^2=10,n(2n+1)(2n-1)/3=2*5*3/3=10,命题成立
假设当n=k（k>2,k为自然数）时命题成立
则当n=k+1时 s(k+1)=s(k)+(2k+1)^2=k(2k+1)(2k-1)/3 + (2k+1)(2k+1)
=(2k+1)/3[(2k^2-k)+(6k+3)]=(2k+1)/3(2k^2+5k+3)=(2k+1)(k+1)(2k+3)/3
=(k+1)[2(k+1)+1][2(k+1)-1]/3 命题同样成立
证毕