Discrete Mathematics with Graph Theory (3rd Edition) 134

# Discrete Mathematics with Graph Theory (3rd Edition) 134 -...

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132 Solutions to Exercises as desired. By the Principle of Mathematical Induction, we conclude that the formula is correct for all n?1. 14. We have a2 = 1 2 a1 = 1 . 1 = 1, a3 = 22a2 = 4 . 1 = 4 = 2 2 , a4 = 32a3 = 3 2 . 22 = 36, a5 = 42a4 = 4 2 .3 2 .22 = 16·9·4 = 576, and a6 = 52a5 = 5 2 .4 2 .3 2 .22 = 25 ·16·9·4 = 14400. It seems that an = (n - 1)2(n - 2)2 ... 3 2 .2 2 = [(n - 1)!]2. So we make the conjecture that an = [(n - 1)!]2 for n ? 1. For n = 1, [(1 - 1)!]2 = 0!2 = 12 = 1, in agreement with the conjectured value of a1. Now we assume the conjecture is true for n = k; that is, we assume that ak = [(k - 1)!]2. We must prove the conjecture is true for n = k + 1; that is, we must prove that ak+1 = [((k + 1) - 1)!]2, that is, that ak+1 = (k!)2. Now ak+1 k2ak = k 2 [(k - 1)!]2 (using the induction hypothesis) [k(k - 1)!]2 = (k!)2 as desired. We conclude that our conjecture is correct for all n ? 1 by the Principle of Mathematical Induction.
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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