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Unformatted text preview: Base Step: n = 1. Then n 3n = 0, and 6  0. Induction step: Assume 6  n 3n . We now need to show 6  ( n +1) 3( n +1), or that 6  n 3 + 3 n 2 + 3 n + 1n1. Simplifying, we get ( n + 1) 3( n + 1) = n 3n + 3 n 2 + 3 n . By the inductive hypothesis, 6  n 3n . Thus, we need to show 6  3 n 2 + 3 n , or 6  3 n ( n + 1). Note that 3  3 n so all we need to show is that n ( n + 1) is divisible by 2. But one of n and n + 1 must be even, so 3 n ( n + 1) must be divisible by 6. 1...
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This note was uploaded on 08/11/2008 for the course MATH 240 taught by Professor Miller during the Fall '08 term at Wisconsin.
 Fall '08
 Miller
 Math

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