1, f (100) =101,which is relatively simple, as others have said.

As for this, there is a wrong solution ahead, don't be misled.

You can find the first few fn(x) to find the pattern.

f 1(x)=( 1+x)/( 1-x)，

F2(x)= f[f 1(x)]=( 1+f 1(x))/( 1-f 1(x))=- 1/x，

f3(x)=( 1-x)/( 1+x)，

F4(x)= x；

……

It can be deduced that f(4k+ 1)(x)=f 1(x),

f(4k+2)(x)=f2(x)，

f(4k+3)(x)=f3(x)，

f(4k+4)(x)= F4(x)；

therefore

f 2006(x)= f(4 * 50 1+2)(x)= F2(x)，

That is, f2006 (x) =-1/x.