(1)已知數列an的前n項和為sn滿足sn=an²;+bn,求證an是等差數列 (2)已知等差數列an的前n項和為sn,求證數列sn/n也成等差數列

(1)已知數列an的前n項和為sn滿足sn=an²;+bn,求證an是等差數列 (2)已知等差數列an的前n項和為sn,求證數列sn/n也成等差數列

Sn=an²;+bn
則:
當n=1時,a1=S1=a+b
當n≥2時,an=Sn-S(n-1)=[an²;+bn]-[a(n-1)²;+b(n-1)]=2an-(a-b),其中n≥2
當n=1時,也滿足上式.
得:an=2an-(a-b)
當n≥2時,an-a(n-1)=2a=常數.
則數列{an}是等差數列.
Sn/n=an+n
則:[S(n+1)/(n+1)]-[Sn/n]=a=常數
數列{Sn/n}也是等差數列.