Solving a Recurrence Relation $3a_{n-1} - 4$ Suppose you have the recurrence relation:
$a_n = 3a_{n-1}-4$
$a_0 = 8$
I am confident I have figured out the pattern, but I am unable to write in a closed form.
$$a_n = 3a_{n-1}-4$$
$$= 3(3a_{n-2} - 4) - 4 = 3 *3 a_{n-2} - 3*4 -4$$
$$= 3(3(3(a_{n-3}-4)-4)-4 = 3*3*3a_{n-3} - 3*3*4 - 3*4 - 4$$
Clearly, the closed form solution must contain $3^n*8$, and a term subtracting $4$ in batches of $3$ with the number of $3$s decreasing. 
Any ideas?
 A: Use $$a_{n}-2=3(a_{n-1}-2).$$
Thus, since $a_n-2$ is a geometric progression, we obtain:  $$a_n-2=(8-2)3^n$$
A: $a_n-3a_{n-1}=-4$
$a_{n-1}-3a_{n-2}=-4$
Subtract to get the homogeneous recurrence
$a_n-4a_{n-1}+3a_{n-2}=0$
with characteristic equation $x^2-4x+3$, which has roots $1$ and $3$.
Therefore, $a_n=b3^n+c1^n$.
Use knowledge of $a_0$ and $a_1$ to solve for $b$ and $c$.
A: You could solve it by finding the characteristic equation.
Given
$$a_n=3a_{n-1}-4$$
you can write this without a constant by subtracting
$$a_{n-1}=3a_{n-2}-4$$
to get
$$a_n=4a_{n-1}-3a_{n-2}$$
Now to find the characteristic equation substitute $r^n$ for $a_n$
$$r^n=4r^{n-1}-3r^{n-2}$$
which is equal to
$$r^{n-2}(r^2-4r+3)=0$$
Solving this equation, r equals 3 or 1 and so the characteristic equation is
$$a_n=p3^n+q$$ 
where p,q are real numbers
Since $a_0=8$ and $a_1=20$
$$8=p+q$$
$$20=3p+q$$
Solving this system of equations shows that $p=6$ and $q=2$
Therefore, the answer is
$$a_n=6*3^n+2$$
A: A small trick for any linear recurrence relation.
Making the problem more general , consider
$$a_n = \alpha\, a_{n-1}+\beta$$
Let $a_n=b_n+k$ and replace
$$b_n+k=\alpha\, b_{n-1} +\alpha\, k+\beta$$ Make
$$k=\alpha\, k+\beta \implies k=\frac \beta {1-\alpha}\implies b_n=\alpha\, b_{n-1} $$ Solve it for $b_n$ (this is simple) and $a_n=b_n+k$.
