问题 选择题
设f1(x)=
2
1+x
,fn+1(x)=f1[fn(x)],且an=
fn(0)-1
fn(0)+2
,则a2013=(  )
A.(
1
2
)2012
B.(
1
2
)
2013
C.(
1
2
)
2014
D.(
1
2
)
2015
答案

由题意可得f1(0)=

2
1+0
=2,

a1=

f1(0)-1
f1(0)+2
=
2-1
2+2
=
1
4

由因为fn+1(x)=f1[fn(x)],

所以an+1=

fn+1(0)-1
fn+1(0)+2
=
f1[fn(0)]-1
f1[fn(0)]+2
=
2
1+fn(0)
-1
2
1+fn(0)
+2
=
1-fn(0)
4+2fn(0)
=-
1
2
fn(0)-1
fn(0)+2
=-
1
2
an

故数列{an}为公比为-

1
2
的等比数列,

故a2013=a1×(-

1
2
)2012=
1
4
×(-
1
2
)2012
=(
1
2
)2014

故选C

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