已知f(x)=2x/1+x,设f1(x)=f(x),fn(x)=f[fn-1(x)],n=1,2,3…
f2(x)=f(f(x))=2f(x)/[1+f(x)]=[2·2x/(1+x)]/[1+2x/(1+x)]=4x/(1+3x)
f3(x)=f(f2(x))=2f2(x)/[1+f2(x)]=[2·4x/(1+3x)]/[1+4x/(1+3x)]=8x/(1+7x)
f4(x)= f(f3(x))=2f3(x)/[1+f3(x)]=[2·8x/(1+7x)]/[1+8x/(1+7x)]=16x/(1+15x)
猜想：fn(x)= f(fn-1(x))=2^nx/[1+(2^n-1)x]
证明：当n=1,n=2时,结论显然成立；
假设当n=k时命题也成立,则当n=k+1时：
fk+1(x)= f(fk(x))=2fk(x)/[1+fk(x)]={2·2^k x/[1+(2^k-1)x)]}/{1+2^k x/[1+(2^k-1)x]}
=[2^(k+1)x]/[1+(2^k-1)x+2^k x]=[2^(k+1)x]/{1+[2^(k+1)-1] x}
所以命题对n=k+1也成立；
由上面的证明可知：命题对一切正整数n都成立.