This is a 2nd degree difference equation and as such it has 2 initial conditions in the general solution.
You can only write the 1st five terms in terms of f(0) and f(1).
Now you can write out the terms this way: f(2)=3f(1)-2f(0) ad so on...or you can write the general solution which is
[f(1)-f(0)]*2n+2[f(0)-f(1)]