Question 354514
{{{f^n(x)}}} = {{{(3^n)x}}}-5({{{1+3+3^2}}}+...+{{{3^(n-1)}}}).
This can be proved by induction:
If {{{n=1}}}, then {{{f^1(x)=f(x)=3x-5*3^0}}}.
If {{{n=2}}}, then {{{f^2(x)=f(x)=3(3x-5)x-5 = (3^2)x-3*5-5=(3^2)x-5*(1+3^1)}}}.
Assume for the purpose of induction that it is true for {{{n = k}}}.  have to show that it is also true for {{{n=k}}}.
Assume this is true:
{{{f^(k-1)(x)}}} = {{{(3^(k-1))x}}}-5({{{1+3+3^2}}}+...+{{{3^(k-2)}}}).
Then {{{f^k(x) = f^(k-1)(f(x))}}},and 
{{{f^k(x)= f^(k-1)(3x-5)}}},
{{{f^k(x) =3^(k-1)(3x-5)}}}-5({{{1+3+3^2}}}+...+{{{3^(k-2)}}}).
{{{f^k(x) =(3^k)x-5*3^(k-1)}}}-5({{{1+3+3^2}}}+...+{{{3^(k-2)}}}) by the distributive property.
{{{f^k(x) =(3^k)x}}}-5({{{1+3+3^2}}}+...+{{{3^(k-2)+3^(k-1)}}}) by associativity.
Therefore by induction the formula for {{{f^n (x)}}} is true for all positive integral n by induction.  Now
{{{f^n(x)}}} = {{{(3^n)x}}}-5({{{1+3+3^2}}}+...+{{{3^(n-1)}}}).  Since 
{{{1+3+3^2}}}+...+{{{3^(n-1)}}}={{{(3^n - 1)/2}}},
then
{{{f^n(x)=(3^n)x-(5/2)(3^n - 1)}}}, or
{{{f^n(x)=(3^n)(x-5/2)+5/2)}}},