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Solve these recurrence relations together with the initial conditions given. a) an=an1+6an2a_n = a_{n−1} + 6a_{n−2} for n2n \geq 2, a0=3,a1=6a_0 = 3, a_1 = 6. b) an=7an110an2a_n = 7a_{n−1} − 10a_{n−2} for n2n \geq 2, a0=2,a1=1.a_0 = 2, a_1 = 1.c) an=6an18an2a_n = 6a_{n−1} − 8a_{n−2} for n2,a0=4,a1=10n \geq 2, a_0 = 4, a_1 = 10. d) an=2an1an2a_n = 2a_{n−1} − a_{n−2} for n2,a0=4,a1=1n \geq 2, a_0 = 4, a_1 = 1. e) an=an2a_n = a_{n−2} for n2,a0=5,a1=1n \geq 2, a_0 = 5, a_1 = −1. f) an=6an19an2a_n = −6a_{n−1} − 9a_{n−2} for n2,a0=3,a1=3n \geq 2, a_0 = 3, a_1 = −3. g) an+2=4an+1+5ana_{n+2} = −4a_{n+1} + 5a_n for n0,a0=2,a1=8n \geq 0, a_0 = 2, a_1 = 8

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(a) Given:

an=an1+6an2,n2a_n=a_{n-1}+6a_{n-2}, n\geq 2

a0=3a_0=3

a1=6a_1=6

Roots Characteristic equation\textbf{Roots Characteristic equation}

Let an=r2a_n=r^2, an1=ra_{n-1}=r and an2=1a_{n-2}=1

r2=r+6r2r6=0Subtract r+6 from each side(r3)(r+2)=0Factorizer3=0 or r+2=0Zero product propertyr=3 or r=2Solve each equation\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*}

Solution recurrence relation\textbf{Solution recurrence relation}

The solution of the recurrence relation is then of the form an=α1r1n+α2r2na_n=\alpha_1r_1^n+\alpha_2r_2^n with r1r_1 and r2r_2 the roots of the characteristic equation.

an=α13n+α2(2)nInitial conditions:3=a0=α1+α26=a1=3α12α2Multiply the first equation by 26=2α1+2α26=3α12α2Add the previous two equations12=5α1125=α1Determine α2 from 3=α1+α2 and α1=125α2=3α1=3125=35\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 an=1253n+35(2)na_n=\dfrac{12}{5}\cdot 3^n+\dfrac{3}{5}\cdot (-2)^n

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