To quote user 9492, "I have been trying to do this question for ages and my lecture notes aren't sufficient to solve the problem, even though the lecturer says they are. I am trying to solve f(n)=6f(n−1)+5. The base case f(0)=2. Apparently the answer should be .(6n)+1. I have no idea how to get to this using the above formula".
Sometimes the best way to solve a problem is just by directly looking at it and attacking the problem without any complicated formulas.
Let $a_n = 6a_{n-1} + 5$. Then:
$a_1 = 6a_0 + 5 = 6a_1 + 5, \therefore a_2 = 6(6a_0 + 5)+5 = 6^{2} a_0 + 6(5) + 5$, $a_3 = 6a_2 + 5 = 6(6^{2} a_0 + 6(5) + 5) + 5 = $ $6^{3}a_0 + 6^2 (5) + 6(5) + 5 = 6^3 a_0 + 5(1+6+6^2)$.
So in general, $a_n = 6^{n} a_0 + 5\frac{1-6^{n}}{1-6}$. As $a_0 = 2$ according to you, we have that
$a_n = 2*6^{n} - 1 + 6^{n} = 3*6^n -1$. How is it that given the recurrence relation you get $6^n +1$?
If $f(0) = 2$ according to you, then $f(1) = (6\times 2) + 5 = 17$. This is clearly not equal to $6^1 + 1$ so there's something wrong with what you have stated
If on the other hand you look at the formula for $a_n$ I have given you (sorry for switching between a_n's and f(n)'s, $a_1 = 3*6^1 - 1 = 17$ which is consistent with the given recurrence relation and initial value of $a_0 = 2$.
Ben