结论：数列{an}的前n项和Sn=an²+bn+c(a≠0),则数列{an}是等差数列的充要条件是c=0.
若{an}为等差数列,则其前n项和Sn=na1+n(n-1)d/2
=nS1+n(n-1)d/2=n(a+b+c)+n(n-1)d/2=an²+bn+c
故d=2[(an²+bn+c)-n(a+b+c)]/n(n-1)
=2(an²-na+c-nc)/n(n-1)=2[na(n-1)-(n-1)c]/n(n-1)
=2(n-1)(na-c)/n(n-1)=常量
必有c=0,因为这时才有d=2a（n-1)n/n(n-1)=2a=常量