已知等比数列{an}的公比为q=-12.(1)若a3=14,求数列{an}的前n项和;(Ⅱ)证明:对任意k∈N+,ak,ak+2,ak+1成等差数列.-数学

题目简介

已知等比数列{an}的公比为q=-12.(1)若a3=14,求数列{an}的前n项和;(Ⅱ)证明:对任意k∈N+,ak,ak+2,ak+1成等差数列.-数学

题目详情

已知等比数列{an}的公比为q=-
1
2

(1)若 a3=
1
4 
,求数列{an}的前n项和;
(Ⅱ)证明:对任意k∈N+,ak,ak+2,ak+1成等差数列.
题型:解答题难度:中档来源:陕西

答案

(1)由 a3=class="stub"1
4
=a1q2,以及q=-class="stub"1
2
可得 a1=1.
∴数列{an}的前n项和 sn=
a1(1-qn)
1-q
=
1×[1-(-class="stub"1
2
)
n
 ]
1+class="stub"1
2
=
2-2•(-class="stub"1
2
)
n
3

(Ⅱ)证明:对任意k∈N+,2ak+2-(ak +ak+1)=2a1 qk+1-a1qk-1-a1qk=a1qk-1(2q2-q-1).
把q=-class="stub"1
2
代入可得2q2-q-1=0,故2ak+2-(ak +ak+1)=0,故 ak,ak+2,ak+1成等差数列.

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