已知等比数列{an}中,an>0,a2=14,S4S2=54,则1a1-1a2+1a3-1a4+…+(-1)n+11an的值为()A.2[1-(-2)n]B.2(1-2n)C.23(1+2n)D.23

题目简介

已知等比数列{an}中,an>0,a2=14,S4S2=54,则1a1-1a2+1a3-1a4+…+(-1)n+11an的值为()A.2[1-(-2)n]B.2(1-2n)C.23(1+2n)D.23

题目详情

已知等比数列{an}中,an>0,a2=
1
4
S4
S2
=
5
4
,则
1
a1
-
1
a2
+
1
a3
-
1
a4
+…+(-1)n+1
1
an
的值为(  )
A.2[1-(-2)n]B.2(1-2nC.
2
3
(1+2n)
D.
2
3
[1-(-2)n]
题型:单选题难度:中档来源:不详

答案

设等比数列{an}的公比为q,∵an>0,∴q>0.经验证q=1不成立.
a2=class="stub"1
4
S4
S2
=class="stub"5
4
,可得
a1q=class="stub"1
4
a1(q4-1)
q-1
a1(q2-1)
q-1
=class="stub"5
4
,及q>0,解得
a1=class="stub"1
2
q=class="stub"1
2

an=a1qn-1=(class="stub"1
2
)n

class="stub"1
a1
-class="stub"1
a2
+class="stub"1
a3
-class="stub"1
a4
+…+(-1)n+1class="stub"1
an

=2-22+23+…+(-1)n+1•2n
=
2[1-(-2)n]
1-(-2)

=class="stub"2
3
[1-(-2)n]

故选D.

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