设数列an的前n项和为Sn,满足an+sn=An^2+Bn+1(A不等于0)an为等差数列,求(B-1)/A
问题描述：
设数列an的前n项和为Sn,满足an+sn=An^2+Bn+1(A不等于0)
an为等差数列,求(B-1)/A
答
设{an}的公差为d,则an=a1+(n-1)d=dn+a1-d
Sn=na1+n(n-1)d/2=d/2*n^2+(a1-d/2)n
an+Sn=d/2*n^2+(a1+d/2)n+a1-d
所以A=d/2
B=a1+d/2
1=a1-d
B-1=(a1+d/2)-(a1-d)=3d/2
(B-1)/A=3