Answer:
A general formula for the sequence is [tex]x_k=\frac{4}{5}4^{k}+\frac{16}{5}(-1)^{k}[/tex].
Step-by-step explanation:
Given a linear homogeneous recurrence of the form [tex]x_{k+2}=c_1x_{k+1}+c_2x_{k}[/tex], with constant coefficients [tex]c_1[/tex], [tex]c_2[/tex] and initial conditions [tex]x_0[/tex], [tex]x_1[/tex] has a characteristic equation given by the formula [tex]x^2=c_1x+c_2[/tex], this equation has a degree 2 and has two roots [tex]r_1[/tex] and [tex]r_2[/tex]. If [tex]r_1\neq r_2[/tex] then [tex]x_n=\alpha_1 r_1^n+\alpha_2 r_2^n[/tex] is a solution of the recurrence relation, where [tex]\alpha _1[/tex] and [tex]\alpha _2[/tex] are the solution of the system
[tex]\left \{ {{x_0=\alpha_1 +\alpha_2 } \atop {x_1=ar_1+br_2}} \right.[/tex]
From the problem we know that [tex]x_0=4[/tex] and [tex]x_1=0[/tex] are the inicial conditions and [tex]c_1=3[/tex] and [tex]c_2=4[/tex]. In order to find the general formula for [tex]x_k[/tex], first we find its characteristic equation
[tex]x^{2}=c_1x+c_2=3x+4[/tex]⇔[tex]x^{2}-3x-4=0[/tex]⇔[tex](x-4)(x-1)=0[/tex]⇔[tex]x=4[/tex] ∨ [tex]x=-1.[/tex]
Now we need to find [tex]\alpha _2[/tex] and [tex]\alpha _2[/tex] using the initial conditions and [tex]r_1=4,[/tex] [tex]r_2=-1[/tex], we need to solve the system of equations
[tex]\left \{ {{ x_0=\alpha _1 +\alpha _2=4 } \atop {x_1=\alpha _1(4)+\alpha _2(-1)=0}} \right,[/tex]
you can add the two equations and find the values [tex]\alpha _1=\frac{4}{5}[/tex] and [tex]\alpha _2=\frac{16}{5}[/tex], so a general formula for [tex]x_k[/tex] is [tex]x_k=\frac{4}{5}4^{k}+\frac{16}{5}(-1)^{k}[/tex].
