This answer is incorrect.

Don't forget 49

I think your calculation is incorrect, the answer I got is \(\frac{1}{\lfloor \sqrt[4]{2016} \rfloor}\)

I don't think I understand this question, isn't the function f_{n} always defined for [0,1]? Am I missing something?

No problem

the equation is 3^{2x+2}+27^{x+1}=36, not 3*(2x+2)+27*(x+1)=36.

you got from 3^{2(x+1)} + 3^{3(x+1))} = 3^{2} * 4 to 1^{x+1} + 3^{x+1} = 4. Check your answer, you divided the right side by 9 and the left side by (3^{2})^{x+1}=9^{x+1}.

I think this can be further simplified:

(2x+x^{2}) /(24x+8x^{2})=(2+x)/(24+8x) (when x!=0)

EP on the right side you divide by 9 and on the left side you divide by 9^{x+1}, you can't do that

You can't divide by 3^{2} like that

I don't think your factoring is correct, the equation

(3^{2})^{x+1} + (3^{3})^{x+1} =(3^{2})^{x+1}*(1+3)

Is not always true (although the only solution is x=0)