f(f(x))= 2(2x+ 1)+ 1 = 4x+3
f[f(f(x))]= 2(4x+3)+ 1 = 8x+7
f { f[f(f(x))]} = 2(8x+7)+ 1 = 16x+ 15
Therefore:
f(f(f(……f(x)……))=
(2^n)x+(2^n)- 1
Let's prove it by mathematical induction.
(1) When n= 1
F(x)=2x+ 1, which is consistent with the known results;
When n=