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) = 18Recursive Part: f (n + 1) = 1/9 * f (n), for n > 0 Consider how to prove the following statement about this given function f using induction. For all nonnegative integers n, f (n) = 2/9(n-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? A. For n = 0, f(n) = f(0) = 18; also 2/9(n-1) = 2/9-1 = 2*9 = 18, so f(n) = 2/9(n-1) for n = 0. B. For n = k, assume f(k) = 2/9(k-1) for some integer k ≥ 0, so f(n) = 2/9(n-1) for n = k. C. For n = k+1, f(k+1) = 2/9(k+1-1) when f(k) = 2/9(k-1) for some integer k ≥ 0, so f(n) = 2/9(n-1) for n = k+1. D. For n = 1, f(n) = f(1) = 1/9*f(0) = 1/9*18 = 2; also 2/9(n-1) = 2/90 = 2, so f(n) = 3n for n = 1. Q2. Which of the following would be a correct Inductive Hypothesis for this proof? A. Prove f(k) = 2/9(k-1) for some integer k ≥ 0. B. Assume f(k+1) = 2/9(k) when f(k) = 2/9(k-1) for some integer k ≥ 0. C. Prove f(k) = 2/9(k-1) for all integers k ≥ 0. D. Assume f(k) = 2/9(k-1) for some integer k ≥ 0. Q3. Which of the following would be a correct completion of the Inductive Step for this proof? A. f(k+1) = 1/9*f(k), which confirms the recursive part of the definition. B. When the inductive hypothesis is true, f(k+1) = 1/9*f(k) = 1/9*2/9(k-1) = 2/9(k-1)+1 = 2/9(k) = 2/9(k+1)-1 C. When the inductive hypothesis is true, f(k+1) = 2/9(k) = 2/9(k+1)-1 = 1/9*2/9(k-1) = 1/9*f(k), which confirms the recursive part of the definition. D. When f(k+1) = 2/9(k+1)-1 = 2/9(k) = 1/9*2/9(k-1); also f(k+1) = 1/9*f(k), so f(k) = 2/9(k-1), confirming the induction hypothesis. Q4. Which of the following would be a correct conclusion for this proof? A. By the principle of mathematical induction, f(n+1) =1/9* f(n) 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) = 2/9(n-1) for all integers n ≥ 0. D. By the principle of mathematical induction, f(k) = 2/9(k-1) implies f(k+1) = 2/9(k) for all integers k ≥ 0.