n=1,
3^2=1/3(4+12+11)成立
假设n=k时,
3^2+...(2k+1)^2=k/3(4k^2+12k+11)
则n=k+1时
左边
=3^2+...+(2k+1)^2+(2k+3)^2
=k/3(4k^2+12k+11)+(2k+3)^2
=1/3(4k^3+24k^2+47k+27)
=(k+1)/3(4k^2+20k+27)
=右边
故成立
=(k+1)/3(4(k+1)^2+12(k+1)+11)