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Let the function f : ℕ → ℝ be defined recursively as follows…

Let the function f : ℕ → ℝ be defined recursively as follows:      Initial Condition:  f (0) = 1 Recursive Part:    f (n + 1) = 3 * f (n), for n ≥ 0 Consider how to prove the following statement about this given function f using induction. f (n) = 3n, for all nonnegative integers n. Select the best response for each question below about how this proof by induction should be done.  Q1. Which is a correct way to prove the Basis Step for this proof?  [Basis] A. For n = 1, f(n) = f(1) = 3*f(0) = 3; also 3n= 31 = 3, so f(n) = 3n for n = 1.B. For n = 0, f(n) = f(0) = 1; also 3n = 30 = 1, so f(n) = 3n for n = 0.C. For n = k+1, f(k+1) = 3(k+1) when f(k) = 3k for some integer k ≥ 0, so f(n) = 3n for n = k+1.D. For n = k, assume f(k) = 3k for some integer k ≥ 0, so f(n) = 3n for n = k. Q2.  Which is a correct way to state the Inductive Hypothesis for this proof?  [InductiveHypothesis] A. Prove f(k) = 3k for some integer k ≥ 0. B. Prove f(k) = 3k for all integers k ≥ 0. C. Assume f(k) = 3k for some integer k ≥ 0. D. Assume f(k+1) = 3(k+1) when f(k) = 3k for some integer k ≥ 0. Q3.  Which is a correct way to complete the Inductive Step for this proof?  [InductiveStep] A. When the inductive hypothesis is true, f(k+1) = 3*f(k) = 3*3k  = 3(k+1). B. f(k+1) = 3*f(k), which confirms the recursive part of the definition. C. When f(k+1) = 3(k+1) = 3*3k; also f(k+1) = 3*f(k), so f(k) = 3k, confirming the induction hypothesis. D. When the inductive hypothesis is true, f(k+1) = 3(k+1) = 3*3k = 3*f(k), which confirms the recursive part of the definition. Q4.  Which is a correct way to state the conclusion for this proof?  [Conclusion] A. By the principle of mathematical induction, f(k) = 3k implies f(k+1) = 3(k+1) for all integers k ≥ 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) = 3*f(n) for all integers n ≥ 0. D. By the principle of mathematical induction, f(n) = 3n for all integers n ≥ 0.