众所周知,对于等差数列{an},若其公差为d,则其前n项和Sn=na1+n(n-1)2d=d2n2+(a1-d2)n.令a=d2,b=a1-d2,则有Sn=an2+bn.反之,若数列{an}的前n项和Sn=an2+bn(a,b为常数),则a1=S1=a+b,当n≥2,an=Sn-Sn-1=[an2+bn]-[a(n-1)2+b(n-1)]=2an+(b-a).又a1=a+b也适合上式,∴an=2an+(b-a)(n It is well known that for the arithmetic sequence {an}, if its tolerance is d, its first n items and Sn=na1+n(n-1)2d=d2n2+(a1-d2)n. Let a=d2, b=a1 -d2, there is Sn = an2 + bn. Conversely, if the first n terms of the series {an} and Sn = an2 + bn (a, b is a constant), then a1 = S1 = a + b, when n ≥ 2, An=Sn-Sn-1=[an2+bn]-[a(n-1)2+b(n-1)]=2an+(ba). In addition, a1=a+b is also suitable for the above formula, ∴an= 2an+(ba)(n