Discrete Mathematics with Graph Theory (3rd Edition) 118

Discrete Mathematics with Graph Theory (3rd Edition) 118 -...

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116 Solutions to Exercises (d) When n = 1, (1 + ~)I = ~, while 1 + ~ = ~ so the result is true. Now suppose that k 2: 1 and the result is true for n = k; that is, suppose (1 + ~)k 2: 1 + ~. We must suppose that the statement is true for n = k + 1; that is, we must prove that (1 + ~ )k+1 2: 1 + ktl. Now 1 k+1 1 k 1 (1 + "2) = (1 +"2) (1 + "2) 2: (1 + ~) ( 1 + ~) (by the induction hypothesis) k 1 k k+1 = 1 + "2 + "2 + 4" > 1 + -2- since k > O. This is what we wanted to show. Thus, by the Principle of Mathematical Induction, the result is true for all natural numbers. (e) When n = 1, (1 + x)n = 1 + x, while 1 + nx = 1 + x. Thus, the result is true for n = 1. Now suppose that k 2: 1 and the result is true for n = k; thus, we suppose that (1 + x)k 2: 1 + kx. We must prove it's true for
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Unformatted text preview: n = k + 1; that is, we must prove that (1 + x)k+1 2: 1 + (k + l)x. Now (1 + x)k+1 (1 + x)k(l + x) > (1 + kx)(l + x) (by the induction hypothesis) 1 + kx + x + kx 2 > 1 + (k + l)x since kx 2 2: 0, and this is what we wanted. So, by the Principle of Mathematical Induction, the given statement is true for all n 2: 1. (f) For n = 2, 1 1 1 1 1 7 13 --+ --+ . .. + -= -+ - = -> -. n + 1 n + 2 2n 3 4 12 24 Now suppose k 2: 2 and the inequality holds for k; that is, suppose that We must prove that the inequality holds for k + 1; that is, we must prove that 1 1 1 1 13 k + 2 + k + 3 + k + 4 + ... + 2k + 2 > 24 Now, 1 1 1 1 1 1 1 = k + 1 + k + 2 + k + 3 + ... + 2k + 2k + 1 + 2k + 2 -k + 1 adding and subtracting -k 1 +1 13 1 1 1 > 24 + 2k + 1 + 2(k + 1) -k + 1...
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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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