Question

# Solve these recurrence relations together with the initial conditions given. a) $a_n = a_{n−1} + 6a_{n−2}$ for $n \geq 2$, $a_0 = 3, a_1 = 6$. b) $a_n = 7a_{n−1} − 10a_{n−2}$ for $n \geq 2$, $a_0 = 2, a_1 = 1.$c) $a_n = 6a_{n−1} − 8a_{n−2}$ for $n \geq 2, a_0 = 4, a_1 = 10$. d) $a_n = 2a_{n−1} − a_{n−2}$ for $n \geq 2, a_0 = 4, a_1 = 1$. e) $a_n = a_{n−2}$ for $n \geq 2, a_0 = 5, a_1 = −1$. f) $a_n = −6a_{n−1} − 9a_{n−2}$ for $n \geq 2, a_0 = 3, a_1 = −3$. g) $a_{n+2} = −4a_{n+1} + 5a_n$ for $n \geq 0, a_0 = 2, a_1 = 8$

Solution

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(a) Given:

$a_n=a_{n-1}+6a_{n-2}, n\geq 2$

$a_0=3$

$a_1=6$

$\textbf{Roots Characteristic equation}$

Let $a_n=r^2$, $a_{n-1}=r$ and $a_{n-2}=1$

\begin{align*} r^2&=r+6 \\ r^2-r-6&=0&\color{#4257b2}\text{Subtract r+6 from each side} \\ (r-3)(r+2)&=0 &\color{#4257b2}\text{Factorize} \\ r-3&=0\text{ or }r+2=0&\color{#4257b2}\text{Zero product property} \\ r&=3\text{ or }r=-2&\color{#4257b2}\text{Solve each equation} \end{align*}

$\textbf{Solution recurrence relation}$

The solution of the recurrence relation is then of the form $a_n=\alpha_1r_1^n+\alpha_2r_2^n$ with $r_1$ and $r_2$ the roots of the characteristic equation.

\begin{align*} a_n&=\alpha_1 \cdot 3^n+\alpha_2\cdot (-2)^n \\ &\color{#4257b2}\text{Initial conditions:} \\ 3=a_0&=\alpha_1+\alpha_2 \\ 6=a_1&=3\alpha_1-2\alpha_2 \\ &\color{#4257b2}\text{Multiply the first equation by 2} \\ 6&=2\alpha_1+2\alpha_2 \\ 6&=3\alpha_1-2\alpha_2 \\ &\color{#4257b2}\text{Add the previous two equations} \\ 12&=5\alpha_1 \\ \color{#c34632} \dfrac{12}{5}&\color{#c34632} =\alpha_1 \\ &\color{#4257b2}\text{Determine }\alpha_2\text{ from }3=\alpha_1+\alpha_2\text{ and }\alpha_1 =\dfrac{12}{5} \\ \color{#c34632} \alpha_2&=3-\alpha_1=3-\dfrac{12}{5}\color{#c34632}=\dfrac{3}{5} \end{align*}

Thus the solution of the recurrence relation is $a_n=\dfrac{12}{5}\cdot 3^n+\dfrac{3}{5}\cdot (-2)^n$

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