数列{an}的前n项和为Sn,存在常数A,B,C,使得an+Sn=An2+Bn+C对任意正整数n都成立.若数列{an}为等差数列,求证:3A-B+C=0.

问题描述:

数列{an}的前n项和为Sn,存在常数A,B,C,使得an+Sn=An2+Bn+C对任意正整数n都成立.若数列{an}为等差数列,求证:3A-B+C=0.

因为{an}为等差数列,设公差为d,由an+Sn=An2+Bn+C,
得a1+(n-1)d+na1+

1
2
n(n-1)d=an+Sn=An2+Bn+C,…(2分)
即(
1
2
d-A)n2+(a1+
d
2
-B)n+(a1-d-C)=0对任意正整数n都成立.…(4分)
所以
1
2
d−A=0
a1+
1
2
d−B=0
a1−d−C=0
,∴A=
1
2
d,B=a1+
1
2
d,C=a1-d,
所以3A-B+C=0.       …(10分)