Discrete Mathematics with Graph Theory (3rd Edition) 132

# Discrete Mathematics with Graph Theory (3rd Edition) 132 -...

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130 Solutions to Exercises 6. [BB] The first six terms are 1,3,7,15,31,63. Our guess is that an = 2 n -1. When n = 1,2 1 -1 = 1, agreeing with al. Now assume that k ~ 1 and the result is true for n = k; that is, assume that ak = 2k -1. We must prove the result is true for n = k+ 1; that is, we must prove that ak+ 1 = 2 k + 1_1. NOwak+l = 2ak + 1 = 2(2k - 1) + 1 (by the induction hypothesis) = 2k+l - 2 + 1 = 2k+l - 1, as required. By the Principle of Mathematical Induction, we conclude that the result is true for all n ~ 1. 7. The first six terms are~, 1 2 3, 6 2 3, 3~3, 15 2 63 and 78 2 13. We guess that an = 5 n tl. Note that when n = 1, the formula gives = ~,agreeing with al. Now assume that ak = 5 k tl for some k ~ 1. We must th 5k+l+1 prove atak+l = -4-. Now ak+l = 5ak-1 __ 5(5 k 4 + 1) -1 by the induction hypothesis 5k+l + 5 5k+l + 1 4 -1= 4 ' as desired. 8. (a) [BB] a2 = l(al + ao) = 1(1 + 1) = 2; a3 = 2(a2 + al) = 2(2 + 1) = 2(3) = 6; a4 = 3(a3 + a2) = 3(6 + 2) = 3(8) = 24; a5 = 4(a4 + a3) = 4(24 + 6) = 4(30) = 120. (b) We guess that an = n! and prove this using the strong form of mathematical induction. Certainly the formula is correct for n = 0 since o!
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