Question

5. For the sequence defined by the recurrence relation ak = 4ak-1 + 6, for each integer k ≥ 1, where ao = 2 a) Use the process of Iteration to find an Explicit formula for the sequence. Simplify. (8) b) Use the Principle of Mathematical Induction to verify the correctness of the formula you found in part 5a. (8)

Answer 1

To find an explicit formula for the sequence defined by the recurrence relation ak = 4ak-1 + 6, for each integer k ≥ 1, where ao = 2, we can use the process of iteration.

Starting with a1 = 2, we can compute the first few terms of the sequence as follows: a1 = 2 a2 = 4a1 + 6 = 14 a3 = 4a2 + 6 = 58 a4 = 4a3 + 6 = 234 a5 = 4a4 + 6 = 938

Looking at these terms, we can make a conjecture for the explicit formula: an = 2 + 4 + 4^2 + ... + 4^(n-2) + 4^(n-1)

We can prove this formula using mathematical induction.

Base Case: For the base case, we let n = 1. Then the formula gives: a1 = 2 = 2 + 4^0 = 2 + 1

This is true, so the base case holds.

Induction Hypothesis: Assume that the formula holds for some arbitrary value k, i.e., ak = 2 + 4 + 4^2 + ... + 4^(k-2) + 4^(k-1)

Induction Step: We want to show that the formula also holds for k+1. That is, ak+1 = 2 + 4 + 4^2 + ... + 4^(k-2) + 4^(k-1) + 4^k

Using the recurrence relation, we have: ak+1 = 4ak + 6 = 4(2 + 4 + 4^2 + ... + 4^(k-2) + 4^(k-1)) + 6 = 2(4^k - 1) + 6 + 4^(k+1) = 2(4^(k+1) - 1) + 2(4 - 1) = 2 + 4 + 4^2 + ... + 4^(k-1) + 4^k + 4^(k+1)

This is exactly the conjectured formula for ak+1. Therefore, by mathematical induction, the formula holds for all positive integers n.

So the explicit formula for the sequence

Step-by-step explanation: