Discrete Mathematics with Graph Theory (3rd Edition) 216

Discrete Mathematics with Graph Theory (3rd Edition) 216 -...

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214 Solutions to Review Exercises (b) This is the problem of putting r = 5 identical marbles into n = 4 boxes. The answer is (4+~-1) = 56. 19. The general power of x in the binomial expansion is X20-k(x-2)k = x 20 - 3k . We have 20 - 3k = 5 for k = 5. The coefficient in question is e50)( _2)5. 20. The general term of the binomial expansion in question is (~) (16x 2 )12-k ( - 2~) k, the exponent of x here being (24 - 2k) - k = 24 - 3k. We want 24 - 3k = -6, so 3k = 30 and k = 10. The term in question is The coefficient is 6; = 3 2 3. 21. There are two statements here. We prove each by mathematical induction. Consider first the inequality, 4 1 3 +2 3 +3 3 + . . ·+(n_1)3 <:. For n = 1, the left side is 0
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Unformatted text preview: 3 , the right side is ~. Since 0 < ~,the statement is true. Now assume that k ~ 1 and the statement is true for n = k; that is, assume We must prove that the statement is true for n = k + 1; that is, We have 1 3 + 2 3 + 3 3 + . .. + k 3 = 1 3 + 2 3 + 3 3 + . .. + (k _1)3 + k 3 k4 < '4 + k 3 , using the induction hypothesis, k4 + 4k 3 k4 + 4k 3 + 6k 2 + 4k + 1 (k + 1)4 4 < 4 4 which is the desired result. By the Principle of Mathematical Induction, inequality (*) holds for all n~1. Next we prove that for all n ~ 1. Since 1 > ~, the inequality is true for n = 1. Now assume k ~ 1 and the inequality is true for n = k; that is,...
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