An=n^2,求Sn=?(要过程)

(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
全加起来，左边中间正负抵消
所以(n+1)^3-1^3=3*(1^2+2^2+……+n^2)+3*(1+2+……+n)+n*1
n^3+3n^2+3n=3*(1^2+2^2+……+n^2)+3n(n+1)/2+n

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

用公式 sn=n（n+1）（2n+1）/6

因An=n^2,Sn=1^2+2^2+...+n^2=n(n+1)(2n+1)/6,加分才有过程