Discrete Mathematics with Graph Theory (3rd Edition) 140

Discrete Mathematics with Graph Theory (3rd Edition) 140 -...

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138 Solutions to Exercises 44. We use the strong form of the Principle of Mathematical Induction. When n = 0, ao = 0 :S 20(0) = 0, so the result holds. Now let k > 0 and assume the result holds for all 0 :S n < k. We wish to prove the result for n = k; that is, we wish to prove that ak :S 20k. Now, ak = aLk/5J + aL3k/5J + k < 20Lk/5j +20L3k/5j +k applying the induction hypothesis to Lk/5j and L3k/5j < 20(~) + 20C:) + k = 4k + 12k + k = 17k < 20k as desired. By the Principle of Mathematical Induction, we conclude that the inequality is true for all n ~ O. 45. Observe that <1>2 = 1 + <1>, hence, <1>3 = <1> + <1>2, <1>4 = <1>2 + <1>3 and, in general, <1>n+l = <1>n + <1>n-l for all n ~ 1. This says that an = <1>n satisfies the Fibonacci recurrence relation. 46. [BB] Rewriting Ik+l = Ik + Ik-l as ik-l = Ik+l - ik, we see that 10 = h - h = 1 - 1 = O. Similarly, 1-1 = 1,1-2 = -1,1-3 = 2,1-4 = -3, 1-5 = 5,1-6 = -8. In general, I-n = (_I)n+l In. 47. [BB] The Fibonacci sequence is the sequence 1,1,2,3,5,8
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