a(n)=a+(n-1)d,
S(n)=na+n(n-1)d/2,
S(2n)=2na+2n(2n-1)d/2,
S(3n)=3na+3n(3n-1)d/2.
S(3n)-S(2n)=na+nd[9n-3-4n+2]/2=na+nd[5n-1]/2,
S(2n)-S(n)=na+dn[4n-2-n+1]/2=na+nd[3n-1]/2,
[S(3n)-S(2n)]-[S(2n)-S(n)]=nd[2n]/2=dn^2,
[S(2n)-S(n)]-[S(n)]=nd[3n-1-n+1]/2=dn^2.
所以,
Sn,S2n-Sn,S3n-S2n,为等差数列,其公差为dn^2.