Answer to Question #302223 in Discrete Mathematics for salomo

Question

Qn 3. Consider a sequence {an}∞

n=0 of real numbers defined as follows:

a0 = 3, a1 = 2, a2 = 12, and

2an+3 − an+2 − 8an+1 + 4an = 0 for all n ≥ 0.

(i) What is a99?

(ii) Find a formula for 2a2n+1 − a2n in terms of n

Answer 1

The characteristic polynomial of this recurrence relation is

$2r^3-r^2-8r+4=0$

$(r-2)(2r^2+3r-2)=0$

$(r-2)(r+2)(2r-1)=0$

The characteristic roots are $r_1=-2, r_2=1/2, r_3=2.$

Hence, the solutions to this recurrence relation are of the form

$a_n=\alpha_1(-2)^n+\alpha_2(\dfrac{1}{2})^n+\alpha_3(2)^n$

To find the constants $\alpha_1,\alpha_2,\alpha_3,$ use the initial conditions. This gives

$a_0=\alpha_1+\alpha_2+\alpha_3=3$

$a_1=-2\alpha_1+\dfrac{1}{2}\alpha_2+2\alpha_3=2$

$a_2=4\alpha_1+\dfrac{1}{4}\alpha_2+4\alpha_3=12$

$\alpha_1=1, \alpha_2=0, \alpha_3=2$

Hence, the unique solution to this recurrence relation and the given initial

conditions is the sequence {an} with

$a_n=(-2)^n+2(2)^n$

(i)

$a_{99}=(-2)^{99}+2(2)^{99}=(2)^{99}$

(ii)

$2a_{2n+1}-a_{2n}=2(-2)^{2n+1}+4(2)^{2n+1}$

$-((-2)^{2n}+2(2)^{2n})$

$=-4(2)^{2n}+8(2)^{2n}-(2)^{2n}-2(2)^{2n}$

$=(2)^{2n}=(4)^{n}$