设f(x)=f1(x)=(1+x)/(x-1) ,f n+1 (x)=f[fn(x)],则f2011(x)=

题目

设f(x)=f1(x)=(1+x)/(x-1) ,f n+1 (x)=f[fn(x)],则f2011(x)=

答案

像这种题目肯定是有一种循环的结果.
f(x)=f1(x)=(1+x)/(x-1)
f2(x)=f[f1(x)]=f((1+x)/(x-1))=x
f3(x)=f(f2(x))=(1+x)/(x-1)
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f2011(x)=(1+x)/(x-1)