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Discrete Mathematics with Graph Theory (3rd Edition) 111

# Discrete Mathematics with Graph Theory (3rd Edition) 111 -...

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Section 5.1 109 (h) For n = 1, 2 n + 3 n - 5 n = 0 is divisible by 6. Now assume that k ::::: 1 and that 2k + 3 k - 5 k is divisible by 6; that is, assume that 5 k = + 3 k + 6N for some integer N. We have 2k+l + 3k+l _ 5 k + 1 = 2(2k) + 3(3 k ) - 5(5 k ) = 2(2k) + 3(3 k ) - 5(2k + 3 k + 6N) using the induction hypothesis, = -3(2k) - 2(3 k ) - 30N = _3(2)(2k-l) - 2(3)(3 k - 1 ) - 30N, which is divisible by 6 because each term is divisible by 6. By the Principle of Mathematical Induction, we conclude that 2 n + 3 n - 5 n is divisible by 6 for all n ::::: 1. (i) For n = 1, 16 n + IOn - 1 = 25 is divisible by 25. Now assume that k ::::: 1 and that k + IOk-l is divisible by 25; that is, assume that k = 1 - 10k + 25N for some integer We have 16k+l + lO(k + 1) - 1 = 16(16 k ) + 10k + 9 = 16(1 - 10k + 25N) + + 9 using the induction hypothesis, = 25 - 150k + 16(25N), which is divisible by 25 because each term is divisible by 25. By the Principle of Mathematical Induction, we conclude that n + IOn - 1 is divisible by 25 for all n ::::: 1. (j) [BB] For n = 1, 2! = 2 is divisible by 2 1 , so the result is true. Now assume that k ::::: 1 and that (2k)! is divisible by 2k. We have [2(k + I)J!
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