Discrete Mathematics with Graph Theory (3rd Edition) 207

Discrete Mathematics with Graph Theory (3rd Edition) 207 -...

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Section 7.7 205 7. (a) 18 terms (b) There are two middle terms (since 17 is odd): C;)x9(3y)8 and C;)x 8 (3y)9. 8. [BB] The required term is C~) (4x)3(5yf = 120(64x 3 )(78125y 7). The coefficient is 600,000,000. 9. [BB] The term in question is (1J)x I3 (3y2)4 = 2380x I3 (81y8). The coefficient is 2380(81) = 192,780. 10. The general term is ekO)x 20 - k ( _2x-2)k = ekO) (_2)k(x 20 - 3k ). We want 20 - 3k = 5, so k = 5 and the coefficient is e50) (_2)5 = -496,128. ( 3)18-k 11. [BB] The general term is C:) -; (x 2 )k = Ck8)318-kx3k-18. We want 3k - 18 = 27, so k = 15 and the coefficient is G~)33 = (816)(27) = 22,032. 12. The general term is (5:) (2x )58-k (:;) k. We want x 58 - k (x- 2 )k = x 25 ; that is, x 58 - 3k = X 25 . Thus, 58 - 3k = 25, 3k = 33, k = 11. The term we seek is and the coefficient of x 25 is - (~~) 2 47 3 11 . 13. We use mathematical induction. If n = 5, 4 5 = 1024 > 625 = 54 so the result is true. Now assume that 4k > k4 for some integer k ~ 5. Then 4 k + 1 =
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Unformatted text preview: 4k > 4 k4 by hypothesis. To prove the desired inequality 4 k + 1 > (k + 1)4 it is, therefore, sufficient to prove that 4 k4 ~ (k + 1)4. We know that (k + 1)4 = k4 + 4k 3 + 6k 2 + 4k + 1. Since 4 < k, 4k 3 < k4 and since 6 < k 2 , 6k 2 < k4. Also, 4k + 1 < 4k + k = 5k ::; kk < k4. Thus, (k + 1)4 < k4 + k4 + k4 + k4 = 4k4 as desired. 14. (a) [BB] Ifn = 0, (1 + v'2)O = 1 = 1 + 0v'2, so Xo = 1, Yo = O. If n = 1, (1 + v'2) I = 1 + v'2, so Xl = 1, YI = 1. If n = 2, (1 + v'2)2 = 1 + 2v'2 + 2 = 3 + 2v'2 so X2 = 3, Y2 = 2. If n = 5, (1 + v'2)5 = 1 + 5v'2 + 1O( v'2)2 + 10( v'2)3 + 5( v'2)4 + (v'2)5 1 + 5v'2 + 20 + 20v'2 + 20 + 4v'2 = 41 + 29v'2 and so X5 = 41, Y5 = 29. (b) We prove that, for n even, Xn 1 + 2(~) + 4(~) + 8(~) + . .. + 2n/2 Yn n + 2(;) + 4(~) + 8(~) + . .. + 2(n-2)/2(n~l) while for n odd, Xn 1 + 2(~) + 4(~) + 8(~) + . .. + 2(n-I)/2(n~l) Yn n + 2(;) + 4(~) + 8(~) + . .. + 2(n-I)/2...
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