数列{an}满足a1=67,an+1=2an(0≤an<12)2an-1(12≤an<1),则a2009=______.-数学

题目简介

数列{an}满足a1=67,an+1=2an(0≤an<12)2an-1(12≤an<1),则a2009=______.-数学

题目详情

数列{an}满足a1=
6
7
,an+1=
2an(0≤an
1
2
)
2an-1(
1
2
an<1)
,则a2009=______.
题型:填空题难度:偏易来源:不详

答案

a1=class="stub"6
7
>class="stub"1
2

a2= class="stub"5
7
> class="stub"1
2

a3=class="stub"3
7

a4=class="stub"6
7

a5=class="stub"5
7
,…
∴数列具有周期性,且周期为3,
a2009=a2=class="stub"5
7

故答案为:class="stub"5
7

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