Chapter 12 Sequences; Induction; the Binomial Theorem

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Chapter 12: Sequences; Induction; the Binomial Theorem 100. 21 We begin with an initial guess of a 0 = 5 . 1⎛ 21⎞ a1 = ⎜5+ ⎟= 4.625 2⎝ 5 ⎠ 1⎛ 21 ⎞ a2 = ⎜4.625 + ⎟≈ 4.58277027 2⎝ 4.625⎠ 1⎛ 21 ⎞ a3 = ⎜4.58277027 + ⎟ 2 ⎝ 4.58277027 ⎠ ≈ 4.582575699 1⎛ 21 ⎞ a4 = ⎜4.582575699 + ⎟ 2 ⎝ 4.582575699 ⎠ ≈ 4.582575695 1⎛ 21 ⎞ a5 = ⎜4.582575695 + ⎟ 2 ⎝ 4.582575695 ⎠ ≈ 4.582575695 101. 89 We begin with an initial guess of a 0 = 9 . 1⎛ 89⎞ a1 = ⎜5 + ⎟≈9.444444444 2⎝ 5 ⎠ 1⎛ 89 ⎞ a2 = ⎜9.444444444 + ⎟ 2 ⎝ 9.444444444 ⎠ ≈ 9.433986928 1⎛ 89 ⎞ a3 = ⎜9.433986928 + ⎟ 2 ⎝ 9.433986928 ⎠ ≈ 9.433981132 1⎛ 89 ⎞ a4 = ⎜9.433981132 + ⎟ 2 ⎝ 9.433981132 ⎠ ≈ 9.433981132 1⎛ 89 ⎞ a5 = ⎜9.433981132 + ⎟ 2 ⎝ 9.433981132 ⎠ ≈ 9.433981132 For both a 5 and the calculator approximation, we obtain 21 ≈ 4.582575695 . For both a 5 and the calculator approximation, we obtain 89 ≈ 9.433981132 . _________________________________________________________________________________________________ 102. To show that 1 2 3 ... ( n 1) ( + 1) n n + + + + − + n = 2 S = 1 + 2 + 3 + ... + n− 1 + n, we can reverse the order to get Let ( ) + S = n+ ( n− ) + ( n− ) 1 2 +...+ 2 + 1, now add these two lines to get [ ] ⎡ ( ) ⎤ ⎡ ( ) ⎤ ⎡( ) ⎤ [ ] 2S = 1+ n + ⎣2 + n− 1 ⎦+ ⎣3+ n− 2 ⎦+ ...... + ⎣ n− 1 + 2⎦+ n+ 1 So we have 2S = [ 1+ n] + [ 1+ n] + [ 1 + n] + .... + [ n+ 1] + [ n+ 1] = n⋅ [ n+ 1] Thus, 2S = n( n+ 1) n⋅ ( n+ 1) S = 103. Answers will vary. 2 1246 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 12.2: Arithmetic Sequences Section 12.2 1. arithmetic 2. False; the sum of the first and last terms equals twice the sum of all the terms divided by the number of terms. 3. d = sn −sn−1 = ( n+ 4) −( n− 1+ 4) = ( n+ 4) − ( n+ 3) = n+ 4−n− 3= 1 The difference between consecutive terms is constant, therefore the sequence is arithmetic. s1 = 1+ 4= 5, s2 = 2+ 4= 6, s3 = 3+ 4= 7, s = 4+ 4= 8 4. 5. 6. 4 d = s −s n n−1 ( ) ( ) = ( n−5) −( n−1− 5) = n−5 − n−6 = n−5− n+ 6= 1 The difference between consecutive terms is constant, therefore the sequence is arithmetic. s1 = 1− 5 =− 4, s2 = 2 − 5 =− 3, s3 = 3 − 5 =−2, s = 4− 5= −1 4 d = an −an−1 = ( 2n−5 ) −(2( n−1) −5) = ( 2n−5) −( 2n−2−5) = 2n−5− 2n+ 7 = 2 The difference between consecutive terms is constant, therefore the sequence is arithmetic. a1 = 21 ⋅ − 5= − 3, a2 = 22 ⋅ − 5= −1, a = 23 ⋅ − 5= 1, a = 24 ⋅ − 5= 3 3 4 d = bn −bn−1 = ( 3n+ 1 ) −(3( n− 1) + 1) = ( 3n+ 1) −( 3n− 3+ 1) = 3n+ 1− 3n+ 2= 3 The difference between consecutive terms is constant, therefore the sequence is arithmetic. b1 = 31 ⋅ + 1= 4, b2 = 32 ⋅ + 1= 7, b = 3⋅ 3+ 1= 10, b = 3⋅ 4+ 1= 13 3 4 7. 8. d = c − c n n−1 ( n) n ( 6 2n) ( 6 2n 2) = 6−2 −(6−2( −1)) = − − − + = 6−2n− 6+ 2n− 2=−2 The difference between consecutive terms is constant, therefore the sequence is arithmetic. c = 6−2⋅ 1= 4, c = 6−2⋅ 2= 2, c 1 2 = 6−2⋅ 3= 0, c = 6−2⋅ 4=−2 3 4 d = d − d n n−1 ( n) n ( 4 2n) ( 4 2n 2) = 4−2 −(4−2( −1)) = − − − + = 4−2n− 4+ 2n− 2=−2 The difference between consecutive terms is constant, therefore the sequence is arithmetic. d = 4−2⋅ 1= 2, d = 4−2⋅ 2= 0, d 1 2 = 4−2⋅ 3= − 2, d = 4−2⋅ 4= −4 3 4 9. d = tn − tn−1 ⎛1 1 ⎞ ⎛1 1 ⎞ = ⎜ − n⎟−⎜ − ( n− 1) ⎟ ⎝2 3 ⎠ ⎝2 3 ⎠ ⎛1 1 ⎞ ⎛1 1 1⎞ = ⎜ − n⎟−⎜ − n+ ⎟ ⎝2 3 ⎠ ⎝2 3 3⎠ 1 1 1 1 1 1 = − n− + n− = − 2 3 2 3 3 3 The difference between consecutive terms is constant, therefore the sequence is arithmetic. 1 1 1 1 1 1 t1 = − ⋅ 1 = , t2 = − ⋅ 2 =− , 2 3 6 2 3 6 1 1 1 1 1 5 t3 = − ⋅ 3 =− , t4 = − ⋅ 4=− 2 3 2 2 3 6 10. d = tn − tn−1 ⎛2 1 ⎞ ⎛2 1 ⎞ = ⎜ + n⎟− ⎜ + ( n− 1) ⎟ ⎝3 4 ⎠ ⎝3 4 ⎠ ⎛2 1 ⎞ ⎛2 1 1⎞ = ⎜ + n⎟− ⎜ + n− ⎟ ⎝3 4 ⎠ ⎝3 4 4⎠ 2 1 2 1 1 1 = + n− − n+ = 3 4 3 4 4 4 The difference between consecutive terms is constant, therefore the sequence is arithmetic. 2 1 11 2 1 7 t1 = + ⋅ 1 = , t2 = + ⋅ 2 = , 3 4 12 3 4 6 2 1 17 2 1 5 t3 = + ⋅ 3 = , t4 = + ⋅ 4= 3 4 12 3 4 3 1247 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Page 3 and 4: Section 12.1: Sequences 34. Answers
Page 5 and 6: Section 12.1: Sequences 16 16 16 2
Page 7 and 8: Section 12.1: Sequences c. Find the
Page 9 and 10: Section 12.1: Sequences (b) ⎛ 0.0
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Page 15 and 16: Section 12.2: Arithmetic Sequences
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Page 19 and 20: Section 12.3: Geometric Sequences;
Page 29 and 30: Section 12.4: Mathematical Inductio
Page 35 and 36: Section 12.5: The Binomial Theorem
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Page 39 and 40: Chapter 12 Review Exercises 50. 8 1
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Page 47 and 48: Chapter 12 Test 2. a1 an an − 1 =
Page 49 and 50: Chapter 12 Test ⎛5⎞ ⎛5⎞ ⎛
Page 51 and 52: Chapter 12 Cumulative Review 3x x 2
Page 53 and 54: Chapter 12 Projects Chapter 12 Proj

Chapter 12 Sequences; Induction; the Binomial Theorem

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