因为(n+1)^3-n^3=(n+1-n)[(n+1)^2+n(n+1)+n^2]=3n^2+3n+1
所以3n^2=(n+1)^3-n^3-3n-1
所以3*1^2+3*2^2+……+3n^2
=[(1+1)^3-1^3-3*1-1]+[(2+1)^3-2^3-3*2-1]+……+[(n+1)^3-n^3-3n-1]
=(n+1)^3-1^3-3*(1+2+3+……+n)-1*n
=(n+1)^3-1-3*n(1+n)/2-n
=n(n+1)(2n+1)/2
所以1^2+2^2+……+n^2=n(n+1)(2n+1)/6