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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 suces 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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This note was uploaded on 01/27/2010 for the course MATH 213 taught by Professor Jack during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JACK
 Math

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