(n+1)^3=n^3+3n^2+3n+1,
所以
(n+1)^3-n^3=3n^2+3n+1
n^3-(n-1)^3=3(n-1)^2+3(n-1)+1
(n-1)^3-(n-2)^3=3(n-2)^2+3(n-2)+1
……
2^3-1^3=3*1^2+3*1+1
以上n个式子叠加得
(n+1)^3-1=3[n^2+(n-1)^2+……+1]+3[n+(n-1)+……+1]+n
n^2+(n-1)^2+……+1
=[(n+1)^3-1-n]/3-[n+(n-1)+……+1]
=(n+1)(n^2+2n)/3-n(n+1)/2
=n(n+1)(2n+1)/6
an=n2前n项和Sn=1²+2²+3²+------+n²=n(n+1)(2n+1)/6