Finding a solution for a recurrence relation when you have a lone constant in the relation

Question

Problem:
$a_n = a_{n-1} - 3, a_0 = -1$

I can't find anything on how to solve for the solution formula when you have a constant in the recurrence relation (in this case the -3). Can someone give me a step by step process on how to solve this? Thanks.

Answer 1

In a case simple like this you can just unfold the recurrence.

$a_0=-1$, $a_1=-1-3$, $a_2=-1-3-3$, $a_3=-1-3-3-3$, and so on. At each step you add a $-3$, so clearly

$$a_n = -1 + n \cdot (-3)= -1-3n.$$

Answer 2

Hint

Suppose that you write $$a_n=b_n+\alpha+\beta n$$ So, the relation write $$b_n+\alpha+\beta n=b_{n-1}+\alpha+\beta (n-1)-3=b_{n-1}+\alpha+\beta n-\beta-3$$ In other words $$b_n=b_{n-1}-\beta-3$$ Selecting $\beta=-3$ makes $$b_n=b_{n-1}$$

I ma sure that you can take from here.