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# Quiz2Spring07 - 3 k ≥ 2 k 2 1 by multiplying by 3 implies...

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Math 213, Section B1, Quiz 2 (SOLUTIONS); Friday, February 2, 2007 1. Using induction, prove that for every integer n 2 3 n 2 n 2 + 1 . Solution. 1) Base of Induction. First, check if the statement 3 n 2 n 2 + 1 holds for n = 2. We have 3 2 = 9 and 2 · 2 2 + 1 = 2 · 4 + 1 = 9. Since 9 9, the required statement does hold for n = 2. 2) Inductive Step. Let k 2 and suppose that 3 k 2 k 2 + 1 is known to hold. We need to derive that 3 k +1 2( k + 1) 2 + 1, that is, 3 k +1 2( k 2 + 2 k + 1) + 1, that is, 3 k +1 2 k 2 + 4 k + 3. The inductive hypothesis
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Unformatted text preview: 3 k ≥ 2 k 2 + 1 by multiplying by 3 implies 3 k +1 ≥ 6 k 2 + 3 . To show that 3 k +1 ≥ 2 k 2 + 4 k + 3 it su±ces to establish that for k ≥ 2 we have 6 k 2 + 3 ≥ 2 k 2 + 4 k + 3. We have: 6 k 2 + 3 ≥ 2 k 2 + 4 k + 3 is equivalent to: 4 k 2 ≥ 4 k by dividing by 4 k > 0, is equivalent to: k ≥ 1 , which holds since by assumption k ≥ 2. Thus for k ≥ 2 we have 6 k 2 + 3 ≥ 2 k 2 + 4 k + 3, as required. 1...
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