Question

(1) Prove by induction that for each n∈N⩾2,n ²(2) Prove by induction that for any n∈N⩾1,8 divides 5 ²ⁿ−1.

(3) Prove by induction that for any n∈N⩾1,n+3<5n ².
(4) Let k∈N. Prove by induction that the k+1-st odd number can be written in the form 2k+1
(5) Let an be the sequence recursively defined by a ₀=1,a ₁=2 and for n⩾2,a ₙ=3a ₙ₋₁−2a ₙ₋₂. Prove by induction that an=2n for all n⩾0.

Answer 1

Using mathematical induction, we proved that the recursively defined sequence aₙ = 3aₙ₋₁−2aₙ₋₂ with a₀=1 and a₁=2 satisfies aₙ=2ⁿ for all n≥0. We verified the base cases for n=0 and n=1 and proved that the recursive step maintains the formula.

Step-by-step explanation:

We will address part (5) of the student's question: proving that for the recursively defined sequence a₀=1, a₁=2 and for n≥2, aₙ=3aₙ₋₁−2aₙ₋₂, it holds that aₙ=2ⁿ for all n≥0.

1. Base case (n=0): We check the base case directly. We have a₀ = 1, and since 2⁰ = 1, the statement holds for n=0.
2. Next, we check the base case for n=1. Here a₁ = 2, which matches 2¹, thus the statement also holds for n=1.
3. Induction Step: Assume for some k ≥ 0, that aₖ = 2ᵏ and aₖ₊₁ = 2ᵏ⁺¹. We need to prove that aₖ₊₂ = 3aₖ₊₁−2aₖ holds the formula aₙ=2ⁿ.
4. We substitute the inductive hypothesis into the recursion: aₖ₊₂ = 3(2ᵏ⁺¹)−2(2ᵏ) = 3(2⋅2ᵏ)−2²ᵏ = 3⋅2⋅2ᵏ−2⋅2ᵏ = 2ᵏ(3⋅2−2) = 2ᵏ⋅2² = 2ᵏ⁺², proving the statement for n=k+2.

By the principle of mathematical induction, the statement aₙ=2ⁿ holds for all natural numbers n≥0.