lec1116 - Example of Induction inequality problem Here is a...

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Example of Induction inequality problem Here is a different inequality that works out, that I did NOT do in lecture. Define a recurrence relation g as follows: g(0) = 2, g(1) = 4, g(n) = 3g(n-1) + 4g(n-2), for all integers n>1. Prove, using strong induction, that g(n) > 4 n , for all n>1. Base Case(s): n=2: g(2) = 3g(1)+4g(0) = 3(4) + 4(2) = 20 > 4 2 n=3: g(3) = 3g(2)+4g(1) = 3(20) + 4(4) = 76 > 4 3 Assume, for some arbitrary n=k, with k>1, that g(k) > 4 k . We must prove under this assumption for n=k+1, g(k+1) > 4 k+1 , g(k+1) = 3g(k) + 4g(k-1) = 3g(k) + 4g(k-1) 3(4 k ) + 4g(k-1), using the inductive hypothesis > 3(4 k ) + 4(4 k-1 ), using the strong inductive hypothesis = 3(4 k ) +4 k = (3+1)(4 k ) = 4(4 k ) = 4 k+1 This proves the inductive step, so we can conclude that g(n) > 4 n , for all n>1.
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Here is the problem I ended up working out in class: Define a recurrence relation g as follows: g(0) = 2, g(1) = 5, g(n) = 5g(n-1) - 4g(n-2), for all integers n>1. Prove, using strong induction, that g(n) = 4 n + 1, for all n>1. Base case(s): n=2: g(2) = 5g(1) – 4g(0) = 5(5) – 4(2) = 17 = 4 2 + 1 n=3: g(3) = 5g(2) – 4g(1) =5(17) – 4(5)= 65 – 4 3 + 1 Inductive hypothesis: Assume for k
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lec1116 - Example of Induction inequality problem Here is a...

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