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Discrete Mathematics with Graph Theory (3rd Edition) 131

# Discrete Mathematics with Graph Theory (3rd Edition) 131 -...

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Section 5.2 Exercises 5.2 1. (a) [BB] a1 = 1; ak+1 = 5ak for k 2': 1. (b) a1 = 5; ak+1 = ak - 2 for k 2': 1. (c) a1 = 4; a2 = 1; ak+2 = ak - ak+1 for k 2': 1. (d) There are two possibilities: a1 = 1; ak+1 = ak + (_1)k+1k for k 2': 1. Also, a1 = 1, a2 = 2, ak = ak-2 + (_1)k for k 2': 3. 2. (a) [BB] 16,8,4,2,1,1,1; (c) 18,9,4,2,1,1,1; (b) 17,8,4,2,1,1,1; (d) 100,50,25,12,6,3,1. 129 3. When n = 1, 3 n - 1 = 30 = 1, which agrees with a1. Now assume the result is true for n = k; that is, assume ak = 3 k - 1 . We must prove the result for n = k + 1; that is, we must prove that ak+1 = 3(k+1)-1 = 3 k . Now ak+1 = 3ak = 3(3 k - 1 ) (by the induction hypothesis) = 3 k , as required. By the Principle of Mathematical Induction, the result is true for all n 2': 1. 4. [BB] When n = 1, the formula becomes (1-1)(1+ 2 1(1 2 +1+ 2 ) = 0, which is a1. Now assume that the formula is correct when n = k. When n = k + 1, (k 1) 3 (k 1)3 (k - 1)(k + 2)(k 2 + k + 2) ak+1 = + + ak = + + 4 4k 3 + 12k2
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