a_n - 5 a_{n-1}=0
is
a_n = a_0 * 5^n.
I have been playing around with Mathematica and digging in Discrete Mathematics books and I am now able to solve equations of the type a_n - c_1 \cdot a_{n-1}=f(n) , for example if a_0=1 the solution of
a_n - 5 a_{n-1}=n^2
is
a_n = \frac{1}{32}(-8n^2-20n-15+47 \cdot 5^n).
Mathematica has a very nice function for solving recurrence equations ( of any order ) which is called RSolve which I only used to verify my own solution. The key to solving recurrence equations of the first order is always finding some sort of sum. Like the sum of the first n integers, which is \frac{1}{2} n(n+1), is in fact the solution of a_n - a_{n-1} = n, with a_0=0.
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