3_Exercise_2

3_Exercise_2 - -4 a k = a k 1 3 For the third inequality we have a k 2 = 5 a k 1-4 a k 1 = 5-4 a k 1< 5-4 4 since a k 1< 4 by hypothesis = 4 By

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Base case: When n = 1 , a 1 = 2 and a 2 = 3 . So 1 a 1 a 2 < 4 . Induction hypothesis: Assume that the statement is true for some k 1 , i.e. 1 a k a k +1 < 4 . Induction step: We want to prove that 1 a k +1 a k +2 < 4 by breaking it into three cases. 1. We already know that 1 a k +1 by the induction hypothesis. 2. For the second inequality, we have a k +2 = 5 a k +1 - 4 a k +1 = 5 - 4 a k +1 5 - 4 a k since a k a k +1 by hypothesis = 5 a k
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Unformatted text preview: -4 a k = a k +1 3. For the third inequality, we have a k +2 = 5 a k +1-4 a k +1 = 5-4 a k +1 < 5-4 4 since a k +1 < 4 by hypothesis = 4 . By combining these three inequalities, we see that 1 ≤ a k +1 ≤ a k +2 < 4 . By mathemati-cal induction, the statement is true for all n ≥ 1 . 1...
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This note was uploaded on 07/24/2011 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.

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