Final answer:
Using strong induction, we prove that the sequence defined by a recursive relationship satisfies a given formula for all n>0. The base case and the inductive step demonstrate the validity of the formula for n=1 and n=2, and then for all n by assuming its truth for all values up to k and showing it holds for k+1.
Step-by-step explanation:
The question involves proving a sequence statement using strong induction. We will prove that the sequence defined by an = 4an-1 - 4an-2 with initial conditions A0 = 6 and A1 = 8, satisfies an = (6-2n)2n for all n > 0.
Base Case: We need to verify the base cases n=1 and n=2.
For n=1, a1 = 8, and the formula gives (6-2(1))21 = 4 * 2 = 8, which is correct.
For n=2, using the given recurrence, a2 = 4a1 - 4a0 = 4 * 8 - 4 * 6 = 32 - 24 = 8, and the formula gives (6-2(2))22 = 2 * 4 = 8, which is also correct.
Induction Hypothesis: Assume that for some k ≥ 2, the formula holds for all n such that 0 < n ≤ k. That is, an = (6-2n)2n for all n ≤ k.
Inductive Step: We need to show the formula holds for n = k+1.
Based on the recurrence relation: ak + 1 = 4ak - 4ak - 1.
Substitute the induction hypothesis into the equation: ak + 1 = 4(6 - 2k)2k - 4(6 - 2(k - 1))2k - 1.
Simplify the equation to show it equals (6 - 2(k+1))2k+1.