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.
答
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+
n(n-1)d=an+Sn=An2+Bn+C,…(2分)1 2
即(
d-A)n2+(a1+1 2
-B)n+(a1-d-C)=0对任意正整数n都成立.…(4分)d 2
所以
,∴A=
d−A=01 2
a1+
d−B=01 2
a1−d−C=0
d,B=a1+1 2
d,C=a1-d,1 2
所以3A-B+C=0. …(10分)