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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*}

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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 $a_n = (-2)^n(2n-1)$.

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$\displaystyle{\large a_{1} = -2\,\quad a_{2} = 12,\quad a_{n} = -4a_{n - 1} - 4a_{n - 2}\,,\quad n \geq 3}$. $$ \sum_{n = 3}^{\infty}a_{n}z^{n} = -4\sum_{n = 3}^{\infty}a_{n - 1}z^{n} - 4\sum_{n = 3}^{\infty}a_{n - 2}z^{n} = -4\sum_{n = 2}^{\infty}a_{n}z^{n + 1} - 4\sum_{n = 1}^{\infty}a_{n}z^{n + 2} $$

$$ \Psi\left(z\right) - a_{1}z - a_{2}z^{2} = -4z\left[\Psi\left(z\right) - a_{1}z\right] - 4z^{2}\Psi\left(z\right) $$

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}

$$ \Psi\left(z\right) = \sum_{n = 1}^{\infty}a_{n}z^{n} = \sum_{n = 1}^{\infty}\left(-1\right)^{n}\left(2n - 1\right)2^{n}z^{n} $$

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% a_{n} \color{#000000}{\large\ =\ } \left(-1\right)^{n}\left(2n - 1\right)2^{n}\,, \qquad\qquad n = 1, 2, 3, \ldots \quad} \\ \\ \hline \end{array} $$

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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}

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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 .

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