Unformatted text preview: 108 Solutions to Exercises (b) If n = 1, n 3 + (n + 1)3 + (n + 2)3 = 1 3 + 2 3 + 3 3 = 36 and 9 I 36. Suppose the result is true for n = k ~ 1; that is, 9 I [k 3 + (k + 1)3 + (k + 2)3]. Then (k + 1)3 + (k + 2)3 + (k + 3)3 = (k+ 1)3 + (k +2)3 + k 3 +9k 2 + 27k+27 = [k 3 + (k+ 1)3 + (k+ 2)3] + [9k 2 + 27k+ 27]. Since 9 divides the first tenn by the induction hypothesis and since 9 clearly divides the second term, 9 divides (k + 1)3 + (k + 2)3 + (k + 3)3, as desired. By the Principle of Mathematical Induction, the result holds for all n ~ 1. (c) [BB] If n = 1, 5 n  1 = 5  1 = 4 is divisible by 4. Suppose the result is true for n = k ~ 1; that is, 4 I (5 k  1). Thus 5 k  1 = 4N for some integer N. Now 5k+l  1 = 5(5 k )  1 = 5(4N + 1)  1 = 20N + 4 = 4(5N + 1), so 4 I (5k+l  1) as desired. By the Principle of Mathematical Induction, the result holds for all n ~ 1. (d) If n = 1, 8 n  3 n = 8  3 = 5 and 5 I 5. Suppose the result is true for n = k ~ 1; that is, 5 I (8 k  3 k ) so that...
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 Summer '10
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 Graph Theory, Natural number, Prime number, 5k, 8 k, 2 5k, 2 7k

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