Discrete Mathematics with Graph Theory (3rd Edition) 115

Discrete Mathematics with Graph Theory (3rd Edition) 115 -...

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Section 5.1 113 (h) When n = 1, the left side is /5 and the right side is i, so the formula holds. Now assume that k 2': 1 and that /5 + 5~9 + 9.~3 + . .. + (4k-3)\4k+1) = 4k~1' We have 1 1 1 1 1·5 + 5·9 + 9·13 + . .. + [4(k + 1) - 3)][4(k + 1) + 1)] 1 1 1 1 = 1·5 + 5·9 + 9·13 + . .. + (4k + 1)(4k + 5) ( 1 1 1 1) 1 = 1·5 + 5·9 + 9·13 + . .. + (4k - 3)(4k + 1) + (4k + 1)(4k + 5) k 1 = 4k + 1 + (4k + 1)(4k + 5) using the induction hypothesis k(4k+5)+1 (4k + 1)(4k + 5) 4k2 + 5k + 1 (4k + 1)(4k + 5) (k + 1) (4k + 1) k + 1 (4k + 1) (4k + 5) 4k + 5 which is the given formula with n = k + 1. By the Principle of Mathematical Induction, we conclude that /5 + i9 + 9.~3 + . .. + (4n-3l(4n+1) = 4n~1 for all n 2': 1. n 7. (a) [BB] 1 + 21 + 22 + 2 3 + . .. + 2 n = L 2k k=O n (b) 12 - 22 + 3 2 - 4
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Unformatted text preview: 2 + . .. + (_1)n-1 n 2 = L(-1)k-1 k 2 k=l n (c) 12 + 3 2 + 52 + . .. + (2n -1)2 = L(2k -1)2 k=l (d) 123 + 234 + 345 + . .. + n(n + l)(n + 2) n = L k(k + l)(k + 2) k=l 1 1 lIn 1 (e) -+-+-+ ... + = L . 12 23 34 n(n + 1) k=l k(k + 1) (Note that the last term on each of the left sides above provides a description of the general term required on the right. See the remarks following the expression (1).) 1 8. (a) [BB] For n = 1, L(i + 1)2i = 2(21) = 4, while n2 n + 1 = 1(22) = 4. Thus, the formula is i=l correct for n = 1. Now suppose that k 2': 1 and the formula is correct for n = k; thus, we suppose that k L(i + 1)2i = k2k+1. i=l We must prove that the formula is correct for n = k + 1; that is, we must prove that k+1 L(i + 1)2i = (k + 1)2k+2. i=l...
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