设等比数列{an}的前n项和为Sn,a4=a1-9,a5,a3,a4成等差数列.(1)求数列{an}的通项公式,(2)证明:对任意k∈N+,Sk+2,Sk,Sk+1成等差数列.-数学

题目简介

设等比数列{an}的前n项和为Sn,a4=a1-9,a5,a3,a4成等差数列.(1)求数列{an}的通项公式,(2)证明:对任意k∈N+,Sk+2,Sk,Sk+1成等差数列.-数学

题目详情

设等比数列{an}的前n项和为Sn,a4=a1-9,a5,a3,a4成等差数列.
(1)求数列{an}的通项公式,
(2)证明:对任意k∈N+,Sk+2,Sk,Sk+1成等差数列.
题型:解答题难度:中档来源:泰安一模

答案

(1)设等比数列{an}的公比为q,
a1q3=a1-9
2a1q2=a1q3+a1q4
,解得
a1=1
q=-2

故数列{an}的通项公式为:an=(-2)n-1,
(2)由(1)可知an=(-2)n-1,
故Sk=
1×[1-(-2)k-1]
1-(-2)
=
1-(-2)k-1
3

所以Sk+1=
1-(-2)k
3
,Sk+2=
1-(-2)k+1
3

∴Sk+1+Sk+2=
1-(-2)k
3
+
1-(-2)k+1
3
=
2-(-2)k-(-2)k+1
3

=
2-(-2)k(1-2)
3
=
2+(-2)k
3

而2Sk=2
1-(-2)k-1
3
=
2-2(-2)k-1
3
=
2+(-2)(-2)k-1
3
=
2+(-2)k
3

故Sk+1+Sk+2=2Sk,即Sk+2,Sk,Sk+1成等差数列

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