# From exercise 30a 2 2 v1 v0 v 2 2as1 52120 624

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Unformatted text preview: (a) sin k =1 6 π k ek = (b) k =0 n n 40. 1 + 3 + 5 + · · · + (2n − 1) = (2k − 1) = 2 k =1 41. (k − 4)2k (b) k =1 39. 2j −2 (c) n k− k =1 k =1 e7 − 1 e−1 1 1 = 2 · n(n + 1) − n = n2 2 For 1 ≤ k ≤ n the k -th L-shaped strip consists of the corner square, a strip above and a strip to the n (2k − 1) = n2 . right for a combined area of 1 + (k − 1) + (k − 1) = 2k − 1, so the total area is k =1 42. n(n + 1) = 465, n2 + n − 930 = 0, (n + 31)(n − 30) = 0, n = 30. 2 43. (35 − 34 ) + (36 − 35 ) + · · · + (317 − 316 ) = 317 − 34 1 2 44. 1− 45. 1 1 −2 22 1 11 − 23 + + 1 1 − 50 51 + ··· + 1 1 −2 32 2 = 50 51 1 1 −2 202 19 + ··· + = 46. (22 − 2) + (23 − 22 ) + · · · + (2101 − 2100 ) = 2101 − 2 47. (a) 399 1 −1=− 202 400 n k =1 1 1 = (2k − 1)(2k + 1) 2 1 = 2 = (b) lim n→+∞ 1 n = 2n + 1 2 n k =1 1 1 − 2k − 1 2k + 1 1− 1 3 + 11 − 35 + 1 n 1 1− = 2 2n + 1 2n + 1 11 − 57 + ··· + 1 1 − 2n − 1 2n + 1 Exercise Set 7.4 n 48. (a) k =1 220 1 = k (k + 1) n k =1 =1− 49. lim n→+∞ 1 2 1− = (b) 1 1 − k k+1 11 − 23 + + 1 1 − n n+1 + ··· + n 1 = n+1 n+1 n =1 n+1 both are valid 50. n n (xi − x) = ¯ 51. 11 − 34 i=1 n n n xi − i=1 n i=1 1 n xi − nx but x = ¯ ¯ x= ¯ i=1 n xi thus i=1 (xi − x) = nx − nx = 0 ¯ ¯ ¯ xi = nx so ¯ i=1 none is valid i=1 n (ak − bk ) = (a1 − b1 ) + (a2 − b2 ) + · · · + (an − bn ) 52. k =1 n n = (a1 + a2 + · · · + an ) − (b1 + b2 + · · · + bn ) = ak − k =1 bk k =1 n (k + 1)4 − k 4 = (n + 1)4 − 1 (telescoping sum), expand the 53. k =1 n (4k 3 + 6k 2 + 4k + 1) = (n + 1)4 − 1, quantity in brackets to get k =1 n n k3 + 6 4 k =1 k2 + 4 k =1 n k3 = k =1 n n 1 = (n + 1)4 − 1 k+ k =1 1 (n + 1)4 − 1 − 6 4 k =1 n n k2 − 4 k =1 n k− k =1 1 k =1 = 1 [(n + 1)4 − 1 − n(n + 1)(2n + 1) − 2n(n + 1) − n] 4 = 1 (n + 1)[(n + 1)3 − n(2n + 1) − 2n − 1] 4 = 1 1 (n + 1)(n3 + n2 ) = n2 (n + 1)2 4 4 m 54. If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2 k =2· k =1 n2 + 2n m(m + 1) = m(m + 1) = ; 2 4 m+1 (2k − 1) if n = 2m + 1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 = k =1 m+1 m+1 k− =2 k =1 1=2· k =1 (m + 1)(m + 2) n2 + 2n + 1 − (m + 1) = (m + 1)2 = 2 4 30 55. 50 · 30 + 49 · 29 + · · · + 22 · 2 + 21 · 1 = 30 k =1 30 k 2 + 20 k (k + 20) = k =1 k= k =1 30 · 31 30 · 31 · 61 + 20 = 18,755 6 2 221 Chapter 7 EXERCISE SET 7.5 1. (a) 1 1 bh = · 4 · 4 = 8 2 2 y A 1 x 1 y (b) The approximation is greater than A, as the rectangles extend beyond the area. 4 f (x∗ )∆x = (4 + 3 + 2 + 1)(1) = 10 k k =1 1 x 1 y (c) The approximation is less than A, as the rectangles lie inside the area. 4 f (x∗ )∆x = (3 + 2 + 1 + 0)(1) = 6 k k =1 1 x 1 y (d) The approximation is equal to A, as can be seen by measuring congruent triangles. 4 f (x∗ )∆x = (3.5 + 2.5 + 1.5 + 0.5)(1) = 8 k k =1 1 x 1 2. (a) 1 1 bh = 4 · 12 = 24 2 2 y 18 9 6 2 3 6 x Exercise Set 7.5 (b) The approximation is less than A, as the rectangles lie inside the area. 222 y 18 4 f (x∗ )∆x = (6 + 9 + 12 + 15)(1) = 42 k k =1 9 6 2 6 2 3 6 2 (c) The approximation is greater than A, as the rectangles extend beyond the area. 3 3 6 x y 18 4 f (x∗ )∆x = (9 + 12 + 15 + 18)(1) = 54 k k =1 9 6 (d) The approximation is equal to A, as can be seen by measuring congruent triangles. x y 18 4 f (x∗ )∆x = (7.5 + 10.5 + 13.5 + 16.5)(1) = 48 k k =1 9 6 3. (a) x∗ = 0, 1, 2, 3, 4 k 5 f (x∗ )∆x = (1 + 2 + 5 + 10 + 17)(1) = 35 k k =1 (b) x∗ = 1, 2, 3, 4, 5 k 5 f (x∗ )∆x = (2 + 5 + 10 + 17 + 26)(1) = 60 k k =1 (c) x∗ = 1/2, 3/2, 5/2, 7/2, 9/2 k 5 f (x∗ )∆x = (5/4 + 13/4 + 29/4 + 53/4 + 85/4)(1) = 185/4 = 46.25 k k =1 4. (a) x∗ = 1, 2, 3, 4, 5 k 5 f (x∗ )∆x = (1 + 8 + 27 + 64 + 125)(1) = 225 k k =1 (b) x∗ = 2, 3, 4, 5, 6 k 5 f (x∗ )∆x = (8 + 27 + 64 + 125 + 216)(1) = 440 k k =1 (c) x∗ = 3/2, 5/2, 7/2, 9/2, 11/2 k 5 f (x∗ )∆x = (27/8 + 125/8 + 343/8 + 729/8 + 1331/8)(1) = 2555/8 = 319.375 k k =1 x 223 5. Chapter 7 (a) x∗ = −π/2, −π/4, 0, π/4 k 4 √ √ √ f (x∗ )∆x = (0 + 1/ 2 + 1 + 1/ 2)(π/4) = (1 + 2)π/4 ≈ 1.896 k k =1 (b) x∗ = −π/4, 0, π/4, π/2 k 4 √ √ √ f (x∗ )∆x = (1/ 2 + 1 + 1/ 2 + 0)(π/4) = (1 + 2)π/4 ≈ 1.896 k k =1 (c) x∗ = −3π/8, −π/8, π/8, 3π/8 k 4 f (x∗ )∆x = cos k k =1 6. (a) √ 3π π π π 3π π π π + cos + cos + cos = π cos cos = π 2 cos /2 ≈ 2.052 8 8 8 84 4 8 8 x∗ = 0, 1, 2, 3, 4 k 4 f (x∗ )∆x = e0 + e1 + e2 + e3 + e4 (1) = (1 − e5 )/(1 − e) = 85.791 k k =1 (b) x∗ = 1, 2, 3, 4, 5 k 4 f (x∗ )∆x = e1 + e2 + e3 + e4 + e5 (1) = e(1 − e5 )/(1 − e) = 233.204 k k =1 (c) x∗ = 1/2, 3/2, 5/2, 7/2, 9/2 k 4 f (x∗ )∆x = e1/2 + e3/2 + e5/2 + e7/2 + e9/2 (1) = e1/2 (1 − e5 )/(1 − e) = 141.446 k k =1 4 f (x∗ )∆x = (2 + 3 + 2 + 1)(1) = 8 k 7. left endpoints: x∗ = 1, 2, 3, 4; k k =1 4 f (x∗ )∆x = (3 + 2 + 1 + 2)(1) = 8 k right endpoints: x∗ = 2, 3, 4, 5; k k =1 8. (a) A = 1 (h1 + h2 )w =...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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