Discrete Mathematics with Graph Theory (3rd Edition) 217

Discrete Mathematics with Graph Theory (3rd Edition) 217 -...

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Chapter 7 With n = k + 1, we have 1 3 + 2 3 + 3 3 + . .. + (k + 1)3 = 1 3 + 2 3 + 3 3 + . .. + k 3 + (k + 1)3 k4 > 4 + (k + 1) 3 , using the induction hypothesis, k4 + 4(k + 1)3 4 k4 + 4k 3 + 12k2 + 12k + 4 4 k4 + 4k 3 + 6k 2 + 4k + 1 (k + 1)4 > 4 4 215 which is the desired result. By the Principle of Mathematical Induction, inequality (**) holds for all n2:1. 22. (a) (~) is the number of ways of choosing k objects from a set of n objects. We must show that (~= ~) + (nk 1 ) counts the same set of possibilities. Let X be one of the n objects. The number of ways to choose k objects, including X, is the number of ways to select k - 1 objects from n - 1, namely, (~=~). The number of ways to choose k objects, excluding X, is the number of ways to select k objects from n - 1, namely, (nkl). Since the two cases-include X, exclude
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Unformatted text preview: X -are mutually exclusive and exhaustive (one must occur), the total number of ways of selecting k objects from n is the sum (~=~) + (nkl), as desired. (b) Use the identity in (a) twice to obtain (~) = (~=D + (n;1) = [(~=D + (~=DJ + [(~=~) + (n;2)J = (~=D +2(~=D + (n;2). Thus (~) -(n; 2) _ (~=~) = 2(~ =~) is even. 23. This is just t (~) (3 n-k )( _1)k = (3-1)n = 2n. (Note that a different expansion for 2 n was found k=O in Problem 34.) 24. If n = 1, m = 2 is even. If n = 2, @ = 6 is even, so assume n 2: 3. Then e:) = en; 1) + e:~11) = (2n -2) + (2n -2) + (2n -2) + (2n -1) n n-1 n-1 n-2 = (2n -2) + (2n -2) + 2 (2n -2) n n-2 n-l so it suffices to prove that (2nn-2) + (2:-=-22) is even, which is the case because these numbers are equal: ( 2n-2) = (2n - 2)! = (2n-2) n n!(n - 2)! n-2'...
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