Suppose that the sequence x0, x1, x2... is defined by x0 = 4, x1 = 0, and xk+2 = 3xk+1+4xk for k≥0. Find a general formula for xk. Be sure to include parentheses where necessary, e.g. to distinguish 1/(2k) from 1/2k

Accepted Solution

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].