Question

已知数列{an}的前n项和Sn=an2+bn(a不等于0),求{an}的通项公式,并证明该数列为等差数列

Answer 1

(1){an}能项公式：n=1,a1=a+b；n>1,an=Sn-Sn-1=an^2+bn-a(n-1)^2-b(n-1)=2an+b-a=>an=2an+b-a(2)证明{an}数列为等差数列：a2-a1=2an*2+b-a-2an-b+a=2an>=2,an-an-1=2an+b-a-2a(n-1)-b+a=2a=>{an}数列为等差数列...