Question

show that the sequence {an} is a solution of the recur- rence relation an = −3an−1 4an−2 if a) an =0. b) an =1. c) an =(−4)n. d) an =2(−4)n 3.

Answer 1

To show that the sequence {an} is a solution of the recurrence relation an = -3an-1 + 4an-2, we substitute the values of an from each option. The values an = 0, an = (-4)^n, and an = 2(-4)^n/3 satisfy the equation, but an = 1 does not.

Step-by-step explanation:

To show that the sequence {an} is a solution of the recurrence relation an = -3an-1 + 4an-2, we need to substitute the values of an from each option and verify if the equation holds true.

1. If an = 0, then -3an-1 + 4an-2 = -3(0) + 4(0) = 0. Therefore, an = 0 is a solution.
2. If an = 1, then -3an-1 + 4an-2 = -3(1) + 4(0) = -3 + 0 = -3. Therefore, an = 1 is not a solution.
3. If an = (-4)^n, then -3an-1 + 4an-2 = -3(-4)^(n-1) + 4(-4)^n-2 = 3(4)^(n-1) + 4(4)^(n-2) = 4^n - 3(4^(n-1)) = 4^n - 3(4^n) = 4^n - 3(4^n) = 0. Therefore, an = (-4)^n is a solution.
4. If an = 2(-4)^n/3, then -3an-1 + 4an-2 = -3(2(-4)^(n-1)/3) + 4(2(-4)^n/3) = -2(-4)^(n-1) + 8(-4)^n/3 = 2(4^(n-1)) + 8(4^n/3) = 2(4^n) + 8(4^n/3) = 2(4^n) + 8(4^n/3) = 0. Therefore, an = 2(-4)^n/3 is a solution.