Question 5 inductive hypothesis (d) inductive step (5) 3. (5.2) In a strong indu

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Question 5 inductive hypothesis (d) inductive step (5) 3. (5.2) In a strong induction proof that P(n) is true for all ne Z', it is shown that vj(PG) Pk + 1). What best describes j? (a) j e z,j 0 (b) j e Z,j 1 ? jeZl sjck (d) j e Z, 1 sjsk (5) 4. (S. 1) Find the error in the following proof of this "theorem": no basis case has been snour "Theorem: Every positive integer equals the next largest positive integer" "Proof: Let P(n) be the proposition "n-n + 1 . To show that P(k) ? P(k + 1), assume that P(k) is true for some k, so that k-k+1. Add 1 to both sides of this equation to obtain k+ 1 k+2, which is P(k + 1). Therefore P(k) P(k + 1) is true. Hence P(n) is true for all positive integers n." (We are looking for a major error, not just that the format is a little sloppy.) (10) 5. (5.3) Consider the mathematical function, f, defined by f(0) 6, f(I) 10, f(2) 16 f(3) 26, f(4) 42, f(5) 68... Give a recursive mathematical function definition (not a computer program function) for the function f(n) as described above, where n e Z, n 2 0. Basis step (show fewest cases necessary): Recursive step (expressed as fn +2)

Explanation / Answer

Hi,
5,
Given the values of the function are
f(0)= 6
f(1)= 10
f(2)= 16
f(3)= 26
f(4)= 42
f(5)= 68
We can see from the above pattern that the function follows the following properties
f(2)=f(0)+f(1)
f(3)=f(1)+f(2)
f(4)=f(2)+f(3)
f(5)=f(3)+f(4)
Therefore with this we can write the recurrence relation like
f(n)=f(n-1)+f(n-2) where f(0)=6 and f(1)=10
we want the recursive relation expressed as f(n+2), so replacing n with n+2 we get
f(n+2)=f(n+1)+f(n)
Thumbs up if this was helpful, otherwise let me know in comments

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