Discrete Mathematics with Graph Theory (3rd Edition) 163

Discrete Mathematics with Graph Theory (3rd Edition) 163 -...

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Chapter 5 161 which is the desired formula for n = k. By the Principle of Mathematical Induction, the formula holds for all n ~ 1. 14. (a) We have al = f(1) = -1, a2 = f(al) = f( -1) = 7, a3 = f(7) = -25, a4 = f(a3) = 103 and a5 = -409. (b) a2 = al implies f(al) = ab so f(J(t)) = f(t). This yields 3 - 4f(t) = 3 - 4t, which gives 3 - 4(3 - 4t) = 3 - 4t or 3 - 12 + 16t = 3 - 4t, 20t = 12, t = ~. Since f(~) = 3 - 4(~) = 3 - 1; = ~, the sequence is constant. (c) The answer is no. al < a2 implies f(t) < f(J(t)), which is 3 - 4t < -9 + 16t, so that t > ~. But a2 < a3 implies -9 + 16t < f( -9 + 16t) = 3 - 4( -9 + 16t) = 39 - 64t, so 80t < 48, that is, t < ~. No such t exists. 15. We have a = 5, d = 4. (a) a32 = 5 + 31(4) = 129; alOO = 5 + 99(4) = 401. (b) Attempting to solve 125 = 5 + (n - 1)4, we get n - 1 = 30. Thus, 125 is the 31st term of this sequence. (c) Attempting to solve 429 = 5 + (n - 1)4, we get n - 1 = 106. Thus, 429 is the 107th term of the sequence. Attempting to solve 1000 = 5
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