Discrete Mathematics with Graph Theory (3rd Edition) 110

Discrete Mathematics with Graph Theory (3rd Edition) 110 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online