Using forward substitution to derive a closed formula for this recurrence relation

Question

I am given the following recurrence relation: $a_n = 2a_{n-1} - 3, a_0 = -1$

I've began forward substituting and gotten to this point:

$a_1 = 2a_{1-1} - 3 = 2(-1)-3 = -5 $

$a_2 = 2a_{2-1} - 3 = 2(2(-1)-3)-3 = -13$

$a_3 = 2a_{3-1} - 3 = 2(2(2(-1)-3)-3)-3 = -29$

...

$\therefore a_n = ?$

I am unsure how to find the closed form for this pattern. Is there some sort of mathematical process I can use to find it? I have been able to do some other problems similar to this with no issue but all of the patterns were pretty simple to figure out just by thinking about it. I apologize if this is a bad question but I am new to this type of mathematics and am trying to learn on my own.

Answer 1

We have $$a_n-3=2(a_{n-1}-3),$$ which says that $a_n-3$ is a geometric progression, which gives $$a_n-3=(-1-3)\cdot2^n$$ or $$a_n=3-2^{n+2}.$$