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