Please help solve the following recurrences Please help with solving the recurrences to get closed form formulas for $a_n$, $b_n$ and $c_n$. Be sure to clearly label the characteristic equation, the roots of the characteristic equation, the general solution and the specific solution.
$a_n=a_{n-1}+2a_{n-2}$ for $n\geq 2$, $a_0=2$ ,$a_1=-5$
$b_n=8b_{n-1}+16b_{n-2}$ for $n\geq 2$, $b_0=-3$, $b_1=-4$
$c_n=3c_{n-1}+6c_{n-2}-24c_{n-3}+24c_{n-4}$ for $n\geq 3$ $c_0=-2$, $c_1=10$, $c_2=-10$, $c_3=54$
This topic is a little hard for me. Do I just plug in $a_0$ and $a_1$ into the equation? Any tips will be helpful. Thank you.
 A: Rewrite the first equation as $a_n - 2 a_{n-1} = -\left( a_{n-1} - 2 a_{n-2} \right)$. This has the form $g_n = -g_{n-1}$ with the general solution $g_n = (-1)^{n-1} g_1$. Therefore 
$$
    a_n - 2 a_{n-1} = (-1)^{n-1} (a_1 - 2 a_0) = 9 (-1)^n
$$
Let $b_n = 2^{-n} a_n$. Multiplying the above recurrence equation with $2^{-n}$ we get
$$\begin{eqnarray}
    b_n - b_{n-1} = 9 (-2)^{-n} \implies b_n &=& b_0 + 9 \sum_{k=1}^n (-2)^{-k} \\ &=& b_0 + 9 \frac{-1/2}{1-(-1/2)} (1 - (-2)^{-n})  \\ &=& -1 + 3 (-1)^n 2^{-n}
\end{eqnarray}
$$
Hence
$$
    a_n = 2^n b_n = 3 (-1)^n - 2^n
$$
Generally, one solves such recurrence equations using trick with linearity. Suppose a particular solution to your recurrence has a form $a_n = q^n$. Substituting this into the recurrence you find $q^{n+2} = q^{n+1} + 2 q^{n}$, implying that $q$ must satisfy $q^2 = q + 2$ which has two roots, $q=2$ and $q=-1$. Since the recurrence equation is linear, general solution is a linear combination of these two solutions:
$$
   a_n = c_1 (-1)^n + c_2 2^{n}
$$
Initial conditions determine $c_1$ and $c_2$.
You are now well equipped to tackle the remaining two problems by yourself.
A: Define formal power series $g \in \mathbb{R}[[x]]$ by
$$g(x) = \sum_{n = 0}^\infty a_n x^n.$$
From the recurrence condition and initial conditions,
$$g(x) - xg(x) - 2x^2g(x) = 2 -7x$$
and solve this equation for $g(x)$ gives
$$g(x) = \frac{2 -7x}{1 - x - 2x^2} = \frac{1}{2x - 1}+ \frac{3}{x + 1}. $$
Therefore, the general term is given by its derivative:
$$ a_n = \frac{1}{n!}\frac{d^n g}{dx^n}(0)  = \frac{(-2)^n}{(2 \cdot 0 - 1)^{n + 1}} + \frac{(-1)^n 3}{(0 + 1)^{n + 1}} = -2^n + (-1)^n 3.$$
Rest of your questions can be done with the same method.
A: $a_n=a_{n-1}+2a_{n-2}$, rewrite it as $a_n-a_{n-1}-2a_{n-2}=0$, then the characteristic equation is $r^2-r-2=0\implies (r-2)(r+1)=0\implies$ $r=2$ or $r=-1$, so the general solution is $a_n=A(2)^n+B(-1)^n$. For the particular solution, since $a_0=2$ and $a_1=-5$, solve the equations:
\begin{equation}
2=A(2)^0+B(-1)^0\\
-5=A(2)^1+B(-1)^1
\end{equation}
then you get $A=-1$ and $B=3$, therefore $a_n=(-1)(2)^n+3(-1)^n$. 
You can solve the rest by using the same method
