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Unformatted text preview: 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 , 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) = 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 . 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 , 1 2 , 3 4 , 7 8 , 15 16 , 31 32 ....
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This note was uploaded on 08/26/2009 for the course MATH 1ABCD taught by Professor All during the Winter '08 term at Foothill College.
 Winter '08
 All

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