Answers of Sydsaeter and Hammond

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8 CHAPTER 3 INTRODUCTORY TOPICS III: MISCELLANEOUS Answers to Even-Numbered Problems 3.1 √ √ √ √ √ √ √ 2. (a) 2 0 + 2 1 + 2 2 + 2 3 + 2 4 = 2(3 + 2 + 3) (b) (x + 0)2 + (x + 2)2 + (x + 4)2 + (x + 6)2 = 4(x 2 + 6x + 14) (c) a1i b2 + a2i b3 + a3i b4 + · · · + ani bn+1 (d) f (x0 ) x0 + f (x1 ) x1 + f (x2 ) x2 + · · · + f (xm ) xm 2·3+3·5+4·7 6 + 15 + 28 49 4. · 100 = · 100 = · 100 ≈ 144.12 1·3+2·5+3·7 3 + 10 + 21 34 6. (a) The total number of people moving from region i. (b) The total number of people moving to region j . 3.2 2. (a + b)6 = a 6 + 6a 5 b + 15a 4 b2 + 20a 3 b3 + 15a 2 b4 + 6ab5 + b6 . (The coefficients are those in the seventh row of Pascal’s triangle in the text.) 4. (a) = 5 3 = 5·4·3 m 5·4·3·2·1 5! 5! = = = . In…show more content…
(i) n3 + n4 = 304 had read A but not B; (ii) n6 = 46; (iii) n8 = 334. (b) We find n(A \ B) = n3 + n4 = 304, n(C \ (A ∪ B)) = n6 = 46, and n( \ (A ∪ B ∪ C)) = n8 = 334. The last equality is a special case of n( \ D) = n( ) − n(D). (The number of persons who are in , but not in D, is the number of persons in all of minus the number of those who are in D.) © Knut Sydsæter and Peter Hammond 2006 10 CHAPTER 3 INTRODUCTORY TOPICS III: MISCELLANEOUS A S1 S4 S7 S3 S8 S6 C S2 S5 B Figure M3.6.8 3.7 2. We prove only (3.2.6); the proof of (3.2.5) is very similar, but slightly easier. For n = 1 the LHS and the RHS of (3.2.6) are both equal to 1. As the induction hypothesis, suppose (3.2.6) is true for n = k, so that k i=1 1 i 3 = 13 + 23 + · · · + k 3 = [ 2 k(k + 1)]2 Then k+1 i=1 k i = 3 i=1 1 1 i 3 + (k + 1)3 = [ 2 k(k + 1)]2 + (k + 1)3 = (k + 1)2 ( 4 k 2 + k + 1) 4. The claim is true for n = 1. As the induction hypothesis, suppose k 3 + (k + 1)3 + (k + 2)3 is divisible by 9. Note that (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 + 9(k 2 + 3k + 3). This is divisible by 9 because the induction hypothesis implies that the sum of the first three terms is divisible by 9, whereas the last term is also obviously divisible by 9. Review Problems for Chapter 3 2. (a) 12 ·4+22 ·5+32 ·6+42 ·7 =

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