Answer: Scroll Down BelowStep-by-step clarification:To substantiate by inference that for all n ≥ 0, Fₙ < 2ⁿ, we will act two steps: base case and introductory step.**Base Case:**For n = 0, we have F₀ = 0. On the other hand, 2⁰ = 1. Since 0 < 1, the base case holds real.**Inductive Step:**Assume that for few dictatorial number k ≥ 0, Fₖ < 2ᵏ (introductory theory).We will explain that for k + 1, Fₖ₊₁ < 2^(k+1).Using the looping description of the Fibonacci series, we have:Fₖ₊₁ = Fₖ + Fₖ₋₁Since k ≥ 0, we experience that k - 1 ≥ -1, and the Fibonacci series is continually non-negative. Therefore, Fₖ and Fₖ₋₁ are two together certain.Using the introductory theory, we experience that Fₖ < 2ᵏ and Fₖ₋₁ < 2ᵏ.Adding these two prejudices together, we receive:Fₖ + Fₖ₋₁ < 2ᵏ + 2ᵏSimplifying the right side, we have:Fₖ + Fₖ₋₁ < 2(2ᵏ)Since Fₖ₊₁ = Fₖ + Fₖ₋₁, we can substitute this in the prejudice:Fₖ₊₁ < 2(2ᵏ)Using the characteristic of exponents, we have:Fₖ₊₁ < 2^(k+1)This does the introductory step.By achieving the base case and the introductory step, we have convinced by initiation that for all n ≥ 0, Fₙ < 2ⁿ.Note: The evidence adopts that the Fibonacci series starts accompanying F₀ = 0 and F₁ = 1, as noticed in the question charge.