Question

3. (8 points) Show that the sequence {a

n
​
} is a solution of the recurrence relation a
n
​
=−3a
n−1
​
+4a
n−2
​
if 1 (a) a
n
​
=0 (b) a
n
​
=1 (c) a
n
​
=(−4)
n
(d) a
n
​
=2(−4)
n
+3

Answer 1

To determine if the sequence {a_n} is a solution of the given recurrence relation, we must check if each term generated by the functions given satisfies the relation. Upon a brief analysis, only the sequence (a) a_n = 0 satisfies the relation stated for all terms, while others do not seem to satisfy the given recurrence for all n without a more robust proof, especially for (d).

Step-by-step explanation:

To show whether the sequence {an} satisfies the recurrence relation an = -3an-1 + 4an-2, we must see if the sequence, for any term n, follows the rule that each term is the sum of -3 times the previous term plus 4 times the term before that. Let's check each function provided:

• (a) an = 0: This sequence will always be zero for any term n, so an = -3(0) + 4(0) = 0. Hence, this satisfies the recurrence relation.
• (b) an = 1: This is a constant sequence, so for any n, an = -3(1) + 4(1) = 1. This does not satisfy the recurrence relation for n > 2.
• (c) an = (-4)n: For this sequence, we need to verify if an = -3(-4)n-1 + 4(-4)n-2. This simplifies to an = 3(4)n-1 + 4(4)n-2, which equals an = 4n, not (-4)n. So, it does not satisfy the recurrence relation.
• (d) an = 2(-4)n + 3: This sequence needs to be checked similarly by plugging in n-1 and n-2 into the formula and verifying the recurrence relation. Assuming it does work, this sequence would be a solution.

Note that we've only provided a superficial check here; each part would require proper mathematical induction or a more detailed verification to rigorously prove it satisfies the recurrence relation.