Chapter 12 Sequences; Induction; the Binomial Theorem

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Chapter 12: Sequences; Induction; the Binomial Theorem n 9. a n =− + 7 2 ⎡ n ⎤ ⎡ ( n −1) ⎤ an − an−1 = ⎢ − + 7 − 7 2 ⎢− + 2 ⎥ ⎣ ⎥ ⎦ ⎣ ⎦ n n−1 =− + 7+ −7 2 2 1 =− 2 The difference between consecutive terms is constant. Therefore, the sequence is arithmetic 1 with common difference d =− and first term 2 1 13 a 1 =− + 7 = . 2 2 The sum of the first n terms of the sequence is given by n Sn = ( a1 + an) 2 n⎛13 ⎛ n ⎞⎞ = 7 2⎜ + 2 ⎜− + 2 ⎟⎟ ⎝ ⎝ ⎠⎠ n⎛27 n⎞ = 2 ⎜ − 2 2 ⎟ ⎝ ⎠ n = ( 27 −n) 4 10. 8 25,10, 4, ,... 5 10 2 = , 4 2 8 5 8 1 2 = , = ⋅ = 25 5 10 5 4 5 4 5 The ratio of consecutive terms is constant. Therefore, the sequence is geometric with common ratio r = 2 5 and first term a 1 = 25 . The sum of the first n terms of the sequence is given by n 2 n 1− ⎛ ⎞ ⎜ 1 5 ⎟ − r Sn = a1 ⋅ = 25⋅ ⎝ ⎠ 1− r 2 1− 5 11. 2n − 3 an = 2n + 1 2n − 3 2( n −1) −3 an − an−1 = − 2n + 1 2( n − 1) + 1 2n−3 2n−5 = − 2n+ 1 2n−1 = ( 2n− 3)( 2n−1) −( 2n− 5)( 2n+ 1) ( 2n+ 1)( 2n−1) 2 2 ( 4n − 8n+ 3) −( 4n −8n−5) = 2 4n −1 8 = 2 4n − 1 The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic. 2n − 3 an = 2n + 1 an−1 2( n −1) −3 2( n − 1) + 1 2n−3 2n−1 = ⋅ 2n+ 1 2n−5 ( 2n− 3)( 2n−1) = 2n+ 1 2n− 5 ( )( ) The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric. −64 1 12. For this geometric series we have r = =− 256 4 1 1 and a 1 = 256 . Since r =− = < 1, the series 4 4 converges and we get a1 256 256 1024 S∞ = 1− r = 1 5 1−( − 4 ) = = 5 4 n ⎡ 2 1 ⎛ ⎞ ⎤ ⎢ − ⎜ 5 ⎟ ⎥ n ⎢ ⎝ ⎠ ⎥ 5⎡ ⎛2⎞ ⎤ = 25⋅ ⎣ ⎦ = 25⋅ ⎢1 − 3 3 ⎜ 5 ⎟ ⎥ ⎢⎣ ⎝ ⎠ ⎥⎦ 5 n 125 ⎡ ⎛2 ⎞ ⎤ = ⎢1 − 3 ⎜ ⎥ 5 ⎟ ⎢⎣ ⎝ ⎠ ⎥⎦ 1282 © 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.
Chapter 12 Test ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎜0⎟ ⎜1⎟ ⎜2⎟ ⎜3⎟ ⎜4⎟ ⎜5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 5 4 3 2 = 243m + 5⋅81m ⋅ 2 + 10⋅27m ⋅ 4 + 10⋅9m ⋅ 8 + 5⋅3m⋅ 16 + 32 5 4 3 2 = 243m + 810m + 1080m + 720m + 240m+ 32 5 5 4 3 2 2 3 4 5 13. ( 3m+ 2) = ( 3m) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m)( 2) + ( 2) 14. First we show that the statement holds for n = 1 . ⎛ 1⎞ ⎜1+ = 1+ 1= 2 1 ⎟ ⎝ ⎠ 1 1 1 1 The equality is true for n = 1 so Condition I holds. Next we assume that ⎜ ⎛ 1+ 1 1 ... 1 n 1 1 ⎟⎜ ⎞⎛ + ⎞⎛ + ⎞ ⎛ + ⎞ = + 2 ⎟⎜ 3 ⎟ ⎜ n ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ is true for some k, and we show the formula then holds for k + 1. We assume that ⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞ ⎜1+ 1 1 ... 1 k 1 1 ⎟⎜ + + + = + 2 ⎟⎜ 3 ⎟ ⎜ k ⎟ . Now we need to show that ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞ ⎜1+ 1 1 ... 1 1 ( k 1) 1 k 2 1 ⎟⎜ + 2 ⎟⎜ + + + = + + = + 3 ⎟ ⎜ k ⎟⎜ k+ 1 ⎟ . ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ We do this as follows: ⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞ ⎡⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎤⎛ 1 ⎞ ⎜1+ 1 1 ... 1 1 1 1 1 ... 1 1 1 ⎟⎜ + 2 ⎟⎜ + 3 ⎟ ⎜ + k ⎟⎜ + k 1 ⎟ = ⎜ + 1 ⎟⎜ + 2 ⎟⎜ + 3 ⎟ ⎜ + k ⎟ ⎜ + k 1 ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ + ⎠ ⎢ ⎥ ⎣⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎦⎝ + ⎠ ⎛ 1 ⎞ = ( k + 1) ⎜1 + (using the induction assumption) k 1 ⎟ ⎝ + ⎠ 1 = ( k+ 1) ⋅ 1+ ( k+ 1) ⋅ k + 1 = k + 1+ 1 = k + 2 Condition II also holds. Thus, formula holds true for all natural numbers. _________________________________________________________________________________________________ 15. The yearly values of the Durango form a geometric sequence with first term a 1 = 31,000 and common ratio r = 0.85 (which represents a 15% loss in value). 1 a 31,000 ( 0.85) n − n = ⋅ The nth term of the sequence represents the value of the Durango at the beginning of the nth year. Since we want to know the value after 10 years, we are looking for the 11 th term of the sequence. That is, the value of the Durango at the beginning of the 11 th year. 11 1 a ( ) 10 11 = a1 ⋅ r − = 31,000 ⋅ 0.85 = 6,103.11 After 10 years, the Durango will be worth $6,103.11. 16. The weights for each set form an arithmetic sequence with first term a 1 = 100 and common difference d = 30 . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence. an = a1 + ( n−1) d a5 = 100 + ( 5 − 1)( 30) = 100 + 4( 30) = 220 n Sn = ( a+ an) 2 S 5 5 = ( 100 + 220) = 5 2 2( 320) = 800 Since he performs 10 repetitions in each set, we multiply the sum by 10 to obtain the total weight lifted. 10( 800) = 8000 The weightlifter will have lifted a total of 8000 pounds after 5 sets. 1283 © 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.
Page 1 and 2: Chapter 12 Sequences; Induction; th
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
Page 11 and 12: Section 12.1: Sequences 97. a. a 1
Page 13 and 14: Section 12.2: Arithmetic Sequences
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
Page 37 and 38: Section 12.5: The Binomial Theorem
Page 39 and 40: Chapter 12 Review Exercises 50. 8 1
Page 41 and 42: Chapter 12 Review Exercises 29. 7
Page 43 and 44: Chapter 12 Review Exercises 51. I:
Page 45 and 46: Chapter 12 Review Exercises c. If t
Page 47: Chapter 12 Test 2. a1 an an − 1 =
Page 51 and 52: Chapter 12 Cumulative Review 3x x 2
Page 53 and 54: Chapter 12 Projects Chapter 12 Proj

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