Solving a Linear Recurrence Relation I made quick progress on this, and then of course got stumped, so here's the problem:
$$a_0 = -1, a_1 = -2, a_n = 4a_{n-1} - 3a_{n-2}$$
So, following the way I was taught to solve this type of problem, I do the following:
Convert it to $$x^n = 4x^{n-1} - 3x^{n-2}$$
Divide by $x^{n-2}$ because it is the smallest exponent, leaving me with:
$x^2 = 4x-3$
Setting equal to zero so I can factor
$x^2-4x+3 = 0$
Factored:
$(x-3)(x-1)$
Giving me roots of 3 and 1, which not being equal, allows me to continue
$$a_n = q_13^n + q_21^n$$
Using the value of $a_0$
Gets me this:
$$a_0 = q_1+q_2 = -1$$
Which is useless I think because getting either $q_1$ or $q_2$ equal to $$-1 - q_{1 or 2}$$
Gets me no where because then, plugging it in gets me:
$$a_1 = -1 - q_2 + q_21^1 = -2$$
Simplified to:
$-1 = -2$
Even I know -1 does not equal -2, I get a similar situation if I try using $a_1$ first, no clue where to go from here, any help much appreciated, thanks guys.
 A: Everything you have done is correct except that you have forgotten to multiply with $3$ in the $a_1$ equation.
More specifically, you have correctly found that $$a_0=-1=q_1\cdot3^0+q_2\cdot1^0 \implies q_2=-1-q_1 \tag1$$ Then you also have that $$a_1=-2=q_1\cdot3^1+q_2\cdot1^1=3q_1-1-q_1 \tag2$$ where the last equality is due to (1). Now equality (2) yields $$-2=2q_1-1 \implies q_1=-\frac12$$ and substituting in (1) you find that $$q_2=-1-q_1=-1-\left(-\frac12\right)=-\frac12$$ In sum $$q_1=q_2=-\frac12$$
A: A solution using generating functions: Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n \qquad a_0 = -1, a_1 = -2
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2}
\end{align}
to get:
$$
\frac{A(z) + 1 + 2 z}{z^2}
  = 4 \frac{A(z) + 1}{z} - 3 A(z)
$$
Solve for $A(z)$ and split into partial fractions:
$$
A(z) = - \frac{1 - 2z}{1 - 4 z + 3 z^2}
     = - \frac{1}{2} \frac{1}{1 - z} - \frac{1}{2} \frac{1}{1 - 3 z}
$$
Those are just two geometric series:
$$
a_n = - \frac{1}{2} - \frac{3^n}{2}
    = - \frac{3^n + 1}{2}
$$
