How to solve a recursive equation I have been given a task to solve the following recursive equation
\begin{align*}
a_1&=-2\\
a_2&= 12\\
a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2, \quad n \geq 3.
\end{align*}
Should I start by rewriting $a_n$ or is there some kind of approach to solve these?
I tried rewriting it to a Quadratic Equation (English isn't my native language, sorry if this is incorrect). Is this the right approach, if so how do I continue?
\begin{align*}
a_n&= -4a_n{}_-{}_1-4a_n{}_-{}_2\\
x^2&= -4x-4\\
0&= x^2 + 4x + 4
\end{align*}
 A: Let $f(x)=\sum_{n=0}^\infty {a_{n+1} \over n!}x^n$. The conditions on $f$ are $f(0)=-2$, $f'(0)=12$, and $4f+4f'+f''=0$. Solving this IVP gives $f(x)=e^{-2x}(8x-2)$. The $n$-th term in the Taylor series of $f$ is ${1 \over n!}(-2)^{n+1}(2n+1) $, so $a_n = (-2)^n(2n-1)$.
A: $\displaystyle{\large a_{1} = -2\,\quad a_{2} = 12,\quad
a_{n} = -4a_{n - 1} - 4a_{n - 2}\,,\quad n \geq 3}$.
$$
\sum_{n = 3}^{\infty}a_{n}z^{n}
=
-4\sum_{n = 3}^{\infty}a_{n - 1}z^{n}
-
4\sum_{n = 3}^{\infty}a_{n - 2}z^{n}
=
-4\sum_{n = 2}^{\infty}a_{n}z^{n + 1}
-
4\sum_{n = 1}^{\infty}a_{n}z^{n + 2}
$$
$$
\Psi\left(z\right) - a_{1}z - a_{2}z^{2}
=
-4z\left[\Psi\left(z\right) - a_{1}z\right]
-
4z^{2}\Psi\left(z\right)
$$
where $\displaystyle{\Psi\left(z\right) \equiv \sum_{n = 1}^{\infty}a_{n}z^{n}.
\quad
z \in {\mathbb C}.\quad \left\vert z\right\vert < {1 \over 2}}$.

\begin{align}
\Psi\left(z\right)
&=
{\left(4a_{1} +a_{2}\right)z^{2} + a_{1}z \over 4z^{2} + 4z + 1}
=
{4z^{2} - 2z \over 4z^{2} + 4z + 1}
=
{4z^{2} - 2z \over \left(1 + 2z\right)^{2}}
=
\left(z - 2z^{2}\right)\,{{\rm d} \over {\rm d} z}\left(1 \over 1 + 2z\right)
\\[3mm]&=
\left(z - 2z^{2}\right)\,{{\rm d} \over {\rm d} z}\sum_{n = 0}^{\infty}\left(-\right)^{n}2^{n}z^{n}
=
\left(z - 2z^{2}\right)\sum_{n = 1}^{\infty}\left(-\right)^{n}2^{n}nz^{n - 1}
\\[3mm]&=
\sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n}nz^{n}
-
\sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n + 1}nz^{n + 1}
=
\sum_{n = 1}^{\infty}\left(-1\right)^{n}2^{n}nz^{n}
-
\sum_{n = 2}^{\infty}\left(-1\right)^{n - 1}2^{n}\left(n - 1\right)z^{n}
\\[3mm]&=
-2z
+
\sum_{n = 2}^{\infty}\left(-1\right)^{n}2^{n}nz^{n}
-
\sum_{n = 2}^{\infty}\left(-1\right)^{n - 1}2^{n}\left(n - 1\right)z^{n}
\\[3mm]&=
-2z
+
\sum_{n = 2}^{\infty}\left(-1\right)^{n}\left(2n - 1\right)2^{n}z^{n}
\end{align}

$$
\Psi\left(z\right)
=
\sum_{n = 1}^{\infty}a_{n}z^{n}
=
\sum_{n = 1}^{\infty}\left(-1\right)^{n}\left(2n - 1\right)2^{n}z^{n}
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
a_{n}
\color{#000000}{\large\ =\ }
\left(-1\right)^{n}\left(2n - 1\right)2^{n}\,,
\qquad\qquad
n = 1, 2, 3, \ldots
\quad}
\\ \\ \hline
\end{array}
$$
A: Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$. Rewrite your recurrence without subtractions in indices:
$$
a_{n + 2} = - 4 a_{n + 1} - 4 a_n
$$
Multiply by $z^n$, add over $n \ge 0$, and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = - 4 \frac{A(z) - a_0}{z} - 4 A(z)
$$
By running the recurrence backwards, you have $a_0 = -1$, and:
$$
A(z) = \frac{2}{(1 + 2 z)^2} - \frac{3}{1 + 2 z}
$$
Remember the generalized binomial theorem, in particular for $m \in \mathbb{N}$:
$$
(1 + u)^{-m}
  = \sum_{k \ge 0} \binom{-m}{k} u^k
  = \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
$$
and you can read off:
\begin{align}
a_n
  &= 2 \cdot (-2)^n \binom{n + 2 - 1}{2 - 1} - 3 \cdot (-2)^n \\
  &= (2 n - 1) \cdot (-2)^n
\end{align}
A: you have to get the descriminent of this trinome :
$$
r^2+4r+4 = 0
$$
which is $r= -2$
and search solution of this form : 
$$
(\alpha n+\beta)(-2)^{n}
$$
for your case :
you have to solve this system:
$$
a_n = (\alpha n+ \beta) (-2)^{n } 
$$
$$
a_0 = -2 
$$
$$
a_1 =  12
$$
it's a simple $2*2$ linear system you can solve it with cramer method or gauss pivot .
