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.-数学

题目简介

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.-数学

题目详情

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.
题型:解答题难度:中档来源:不详

答案

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

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