Question

Take into account the starting conditions an = an-1 + 4an-2 , and a0 = 3, a1 = 6, which form the recursive function an as specified by the recurrence equation. To demonstrate that 3 | an ( for every n ∈ N ) we would need to...

(a). For our Basic Case, we would have to demonstrate
3 | a₀

Answer 1

To demonstrate that 3 divides every term of the given sequence, we can use mathematical induction to prove that 3 divides both the given base case and the recursive step.

Step-by-step explanation:

To demonstrate that 3 divides every term of the sequence, we will use mathematical induction. First, we will prove the base case, which is to show that 3 divides the first term, a0. In this case, a0 = 3, which is divisible by 3. Therefore, 3 divides a0.

Next, we assume that 3 divides ak-1 and ak-2 for some positive integer k. Now, we need to prove that 3 divides ak as well.

Using the given recursive formula, ak = ak-1 + 4ak-2. Since we assume that 3 divides both ak-1 and ak-2, we can express them as multiples of 3: ak-1 = 3n and ak-2 = 3m, where n and m are integers.

Substituting these expressions into the formula, we get ak = 3n + 4(3m) = 3(n + 4m). Therefore, ak is a multiple of 3, which means that 3 divides ak.

Since we have proven the base case and the recursive step, we conclude that 3 divides every term of the sequence, an, for every positive integer n.