數學歸納法(3)

Notice that
f1=1/(1-x)
f2=(x-1)/x
f3=x
f4=1/(1-x)
we can deduce that
fn
=1/(1-x) [when n=3m-2]
=(x-1)/x [when n=3m-1]
=x [when n=3m]
for m>=1...(*)
Let P(n) be the statement
The formula of fn has the form given by (*)
when n=1,2,3 we have proved that P(n) is true
Assume that P(k) is true
when n=k+1
if k= 3m-2
Then
fk+1
=f3m-1
=f(fk)
=f(1/(1-x))
=(x-1)/x
which is the form given by (*)
if k= 3m-1
Then
fk+1
=f3m
=f(fk)
=f((x-1)/x)
=x
which is the form given by (*)
if k= 3m
Then
fk+1
=f3(m+1)-1
=f3m'-1
=f(fk)
=f(x)
=1/(1-x)
which is the form given by (*)
So P(k+1) is true
By the mathematical induction, for all positive integer n, P(n) is true