Let the functiоn f : ℕ → ℝ be defined recursively аs fоllоws: Initiаl Condition: f (0) = 0Recursive Pаrt: f (n) = (2 * f (n-1)) + 1, for all n > 0 Consider how to prove the following statement about this given function f using induction. For all nonnegative integers n, f (n) = 2n- 1 Select the best response for each question below about how this proof by induction should be done. Q1. Which of the following would be a correct Basis step for this proof? [Basis] A. For n = k, assume f(k) = 2k - 1 for some integer k ≥ 0, so f(n) = 2n - 1 for n = k. B. For n = 1, f(n) = f(1) = 2*f(0) +1 = 1; also 2n - 1 = 21 – 1 = 2 – 1 = 1, so f(n) = 2n - 1 for n = 1. C. For n = k+1, f(k+1) = 2(k+1) - 1 when f(k) = 2k - 1 for some integer k ≥ 0, so f(n) = 2n - 1 for n = k+1. D. For n = 0, f(n) = f(0) = 0; also 2n - 1 = 20 – 1 = 1 – 1 = 0, so f(n) = 2n - 1 for n = 0. Q2. Which of the following would be a correct Inductive Hypothesis for this proof? [InductiveHypothesis] A. Assume f(k+1) = 2(k+1) - 1 when f(k) = 2k - 1 for some integer k ≥ 0. B. Assume f(k) = 2k - 1 for some integer k ≥ 0. C. Prove f(k) = 2k - 1 for some integer k ≥ 0. D. Prove f(k) = 2k - 1 for all integers k ≥ 0. Q3. Which of the following would be a correct completion of the Inductive Step for this proof? [InductiveStep] A. f(k+1) = 2*f(k) + 1, which confirms the recursive part of the definition. B. When f(k+1) = (2(k+1) - 1) = (2(k+1) – 2) + 1 = 2*(2k - 1) + 1; also f(k+1) = 2*f(k) + 1, so f(k) = (2k - 1), confirming the induction hypothesis. C. When the inductive hypothesis is true, f(k+1) = 2*f(k) + 1 = 2*(2k - 1) + 1 = (2(k+1) – 2) + 1 = (2(k+1) - 1). D. When the inductive hypothesis is true, f(k+1) = (2(k+1) - 1) = (2(k+1) – 2) + 1 = 2*(2k - 1) + 1 = 2*f(k) + 1, which confirms the recursive part of the definition. Q4. Which of the following would be a correct conclusion for this proof? [Conclusion] A. By the principle of mathematical induction, f(n) = (2n – 1) for all integers n ≥ 0. B. By the principle of mathematical induction, f(k) = f(k+1) for all integers k ≥ 0. C. By the principle of mathematical induction, f(n+1) = (2*f(k)) + 1 for all integers n ≥ 0. D. By the principle of mathematical induction, f(k) = (2k – 1) implies f(k+1) = (2(k+1) – 1) for all integers k ≥ 0.
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