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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 Spring '08
 JACK
 Math, Logic, Inductive Reasoning, Natural number, Section B1

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