数列{an}中a1=-1\2 an+1=1\1-an（n=1,2••）a2004=?通项公式
人气：302 ℃ 时间：2020-06-29 01:52:29
解答
a1=-1/2,
a(n+1)=1/(1-an)
a2=1/[1-(-1/2)]
=1/(1+1/2)
=1/(3/2)
=2/3
a3=1/(1-a2)
=1/(1-2/3)
=1/(1/3)
=3
a4=1/(1-a3)
=1/(1-3)
=-1/2
an是关于-1/2,2/3,3循环
2004/3=668
所以a2004=3
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