设f1(x)=2/(1+x),fn+1(x)=f1[fn(x)]设f1(x)=2/(1+x),设fn+1(x)=f1〔fn(x)〕,an=〔fn(0)-1〕/〔fn(0)+2〕,n∈N*,求数列{an}的2009项

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设f1(x)=2/(1+x),fn+1(x)=f1[fn(x)]设f1(x)=2/(1+x),设fn+1(x)=f1〔fn(x)〕,an=〔fn(0)-1〕/〔fn(0)+2〕,n∈N*,求数列{an}的2009项
设f1(x)=2/(1+x),fn+1(x)=f1[fn(x)]
设f1(x)=2/(1+x),设fn+1(x)=f1〔fn(x)〕,an=〔fn(0)-1〕/〔fn(0)+2〕,n∈N*,求数列{an}的2009项

设f1(x)=2/(1+x),fn+1(x)=f1[fn(x)]设f1(x)=2/(1+x),设fn+1(x)=f1〔fn(x)〕,an=〔fn(0)-1〕/〔fn(0)+2〕,n∈N*,求数列{an}的2009项
因为n∈N*,
所以当n=1时,根据条件得:f2(x)=2/(1+f1(x))=(2+2x)/(3+x); a1=1/4
n=2时,根据条件得:a2=-1/8
因为an=〔fn(0)-1〕/〔fn(0)+2〕=1-3/(fn(0)+2)
所以an+1=〔fn+1(0)-1〕/〔fn+1(0)+2〕=-1/2+3/(2fn(0)=4);
所以an+1=-1/2an
所以an=(-1/2)(n+1) 后面(n+1)是次方
所以a2009=(-1/2)2010 后面2010是次方

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