{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch09 - 9.1 SOLUTIONS 633 CHAPTER NINE Solutions for Section...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
9.1 SOLUTIONS 633 CHAPTER NINE Solutions for Section 9.1 Exercises 1. The first term is 2 1 + 1 = 3 . The second term is 2 2 + 1 = 5 . The third term is 2 3 + 1 = 9 , the fourth is 2 4 + 1 = 17 , and the fifth is 2 5 + 1 = 33 . The first five terms are 3 , 5 , 9 , 17 , 33 . 2. The first term is 1 + ( - 1) 1 = 1 - 1 = 0 . The second term is 2 + ( - 1) 2 = 2 + 1 = 3 . The third term is 3 - 1 = 2 and the fourth is 4 + 1 = 5 . The first five terms are 0 , 3 , 2 , 5 , 4 . 3. The first term is 2 · 1 / (2 · 1 + 1) = 2 / 3 . The second term is 2 · 2 / (2 · 2 + 1) = 4 / 5 . The first five terms are 2 / 3 , 4 / 5 , 6 / 7 , 8 / 9 , 10 / 11 . 4. The first term is ( - 1) 1 (1 / 2) 1 = - 1 / 2 . The second term is ( - 1) 2 (1 / 2) 2 = 1 / 4 . The first five terms are - 1 / 2 , 1 / 4 , - 1 / 8 , 1 / 16 , - 1 / 32 . 5. The first term is ( - 1) 2 (1 / 2) 0 = 1 . The second term is ( - 1) 3 (1 / 2) 1 = - 1 / 2 . The first five terms are 1 , - 1 / 2 , 1 / 4 , - 1 / 8 , 1 / 16 . 6. The first term is (1 - 1 / (1 + 1)) (1+1) = (1 / 2) 2 . The second term is (1 - 1 / 3) 3 = (2 / 3) 3 . The first five terms are (1 / 2) 2 , (2 / 3) 3 , (3 / 4) 4 , (4 / 5) 5 , (5 / 6) 6 . 7. The terms look like powers of 2 so we guess s n = 2 n . This makes the first term 2 1 = 2 rather than 4 . We try instead s n = 2 n +1 . If we now check, we get the terms 4 , 8 , 16 , 32 , 64 , . . . , which is right. 8. We compare with positive powers of 2 , which are 2 , 4 , 8 , 16 , 32 , . . . . Each term is one less, so we take s n = 2 n - 1 . 9. We observe that if we subtract 1 from each term of the sequence, we get 1 , 4 , 9 , 16 , 25 , . . . , namely the squares 1 2 , 2 2 , 3 2 , 4 2 , 5 2 , . . . . Thus s n = n 2 + 1 . 10. First notice that s n = 2 n - 1 is a formula for the general term of the sequence 1 , 3 , 5 , 7 , 9 , . . . . To obtain the alternating signs in the original sequence, we try multiplying by ( - 1) n . However, checking ( - 1) n (2 n - 1) for n = 1 , 2 , 3 , . . . gives - 1 , 3 , - 5 , 7 , - 9 , . . . . To get the correct signs, we multiply by ( - 1) n +1 and take s n = ( - 1) n +1 (2 n - 1) . 11. The numerator is n . The denominator is then 2 n + 1 , so s n = n/ (2 n + 1) . 12. The denominators are the even numbers, so we try s n = 1 / (2 n ) . To get the signs to alternate, we try multiplying by ( - 1) n . That gives - 1 / 2 , 1 / 4 , - 1 / 6 , 1 / 8 , - 1 / 10 , . . . , so we multiply by ( - 1) n +1 instead. Thus s n = ( - 1) n +1 / (2 n ) . 13. We have s 2 = s 1 + 2 = 3 and s 3 = s 2 + 3 = 6 . Continuing, we get 1 , 3 , 6 , 10 , 15 , 21 . 14. We have s 2 = 2 s 1 + 3 = 2 · 1 + 3 = 5 and s 3 = 2 s 2 + 3 = 2 · 5 + 3 = 13 . Continuing, we get 1 , 5 , 13 , 29 , 61 , 125 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
634 Chapter Nine /SOLUTIONS 15. We have s 2 = s 1 + 1 / 2 = 0 + (1 / 2) 1 = 1 / 2 and s 3 = s 2 + (1 / 2) 2 = 1 / 2 + 1 / 4 = 3 / 4 . Continuing, we get 0 , 1 2 , 3 4 , 7 8 , 15 16 , 31 32 . 16. We have s 3 = s 2 + 2 s 1 = 5 + 2 · 1 = 7 and s 4 = s 3 + 2 s 2 = 7 + 2 · 5 = 17 . Continuing, we get 1 , 5 , 7 , 17 , 31 , 65 . Problems 17. (a) matches (IV), since the sequence increases toward 1 . (b) matches (III), since the odd terms increase toward 1 and the even terms decrease toward 1 . (c) matches (II), since the sequence decreases toward 0 . (d) matches (I), since the sequence decreases toward 1 . 18. (a) matches (II), since the sequence increases toward 2 . (b) matches (III), since the even terms decrease toward 2 and odd terms decrease toward - 2 . (c) matches (IV), since the even terms decrease toward 2 and odd terms increase toward 2 . (d) matches (I), since the sequence decreases toward 2 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern