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Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-12 JWDD027-Salas-v1 December 2, 2006 16:55 632 SECTION 12.1 CHAPTER 12 SECTION 12.1 1. 1 + 4 + 7 = 12 2. 2 + 5 + 8 + 11 = 26 3. 1 + 2 + 4 + 8 = 15 4. 1 2 + 1 4 + 1 8 + 1 16 = 15 16 5. 1 2 + 4 8 = 5 6. 2 4 + 8 16 = 10 7. 1 3 + 1 9 + 1 27 = 13 27 8. 1 6 + 1 24 1 120 = 2 15 9. 1 + 1 4 + 1 16 + 1 64 = 85 64 10. 1 1 4 + 1 16 1 64 = 51 64 11. 11 n =1 (2 n 1) 12. 10 k =1 ( 1) k +1 (2 k 1) 13. 35 k =1 k ( k + 1) 14. n k =1 m k Δ x k 15. n k =1 M k Δ x k 16. n k =1 f ( x k x k 17. 10 k =3 1 2 k , 7 i =0 1 2 i +3 18. 10 k =3 k k k ! , 7 i =0 ( i + 3) i +3 ( i + 3)! 19. 10 k =3 ( 1) k +1 k k + 1 , 7 i =0 ( 1) i i + 3 i + 4 20. 10 k =3 1 2 k 3 , 7 i =0 1 2 i + 3 21. Set k = n + 3 . Then n = 1 when k = 2 and n = 7 when k = 10 . 10 k =2 k k 2 + 1 = 7 n = 1 n + 3 ( n + 3) 2 + 1 = 7 n = 1 n + 3 n 2 + 6 n + 10 22. 12 n =2 ( 1) n n 1 = 11 k =1 ( 1) k +1 k + 1 1 = 11 k =1 ( 1) k +1 k 23. Set k = n 3 . Then n = 7 when k = 4 and n = 28 when k = 25 . 25 k =4 1 k 2 9 = 28 n =7 1 ( n 3) 2 9 = 28 n =7 1 n 2 6 n 24. 15 k =0 3 2 k k ! = 13 n = 2 3 2( n +2) ( n + 2)! = 81 13 n = 2 3 2 n ( n + 2)! 25. 0 .a 1 a 2 · · · a n = a 1 10 + a 2 10 2 + · · · + a n 10 n = n k =1 a k 10 k
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-12 JWDD027-Salas-v1 December 2, 2006 16:55 SECTION 12.2 633 26. n k =1 1 k = 1 1 + 1 2 + 1 3 + · · · + 1 n 1 n + 1 n + 1 n + · · · + 1 n = n n = n . 27. 50 k =0 1 4 k = 1 . 3333 · · · 28. 50 k =1 1 k 2 = 1 . 62513 29. 50 k =0 1 k ! = 2 . 71828 · · · 30. 50 k =0 2 3 k = 3 SECTION 12.2 1. 1 2 ; s n = 1 2 1 1 · 2 + 1 2 · 3 + · · · + 1 ( n )( n + 1) = 1 2 1 1 2 + 1 2 1 3 + · · · + 1 n 1 n + 1 = 1 2 1 1 n + 1 1 2 2. 1 2 ; k =3 1 k 2 k = k =3 1 k 1 1 k = lim n →∞ 1 2 1 n = 1 2 3. 11 18 ; s n = 1 1 · 4 + 1 2 · 5 + · · · + 1 n ( n + 3) = 1 3 1 1 4 + 1 2 1 5 + · · · + 1 n 1 n + 3 = 1 3 1 + 1 2 + 1 3 1 n + 1 1 n + 2 1 n + 3 1 3 1 + 1 2 + 1 3 = 11 18 4. 3 4 ; k =0 1 ( k + 1)( k + 3) = 1 2 k =0 1 k + 1 1 k + 3 = 1 2 lim n →∞ 1 + 1 2 1 n + 2 1 n + 3 = 3 4 5. 10 3 ; k =0 3 10 k = 3 k =0 1 10 k = 3 1 1 1 / 10 = 30 9 = 10 3 6. 5 6 ; k =0 ( 1) k 5 k = k =0 1 5 k = 1 1 + 1 5 = 5 6 7. 3 2 ; k =0 1 2 k 3 k = k =0 1 3 k k =0 2 3 k = 1 1 1 / 3 1 1 2 / 3 = 3 2 3 = 3 2 8. 1 4 ; k =0 1 2 k +3 = 1 8 k =0 1 2 k = 1 8 · 1 1 1 2 = 1 4 9. 24; geometric series with a = 8 and r = 2 3 , sum = a 1 r = 24
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-12 JWDD027-Salas-v1 December 2, 2006 16:55 634 SECTION 12.2 10. 3 15 , 616 ; k =2 3 k 1 4 3 k +1 = k =0 3 k +1 4 3 k +7 = 3 4 7 k =0 3 4 3 k = 3 4 7 · 1 1 3 4 3 = 3 15 , 616 11. Let x = 0 . a 1 a 2 · · · a n a 1 a 2 · · · a n · · · . Then x = k =1 a 1 a 2 · · · a n (10 n ) k = a 1 a 2 · · · a n k =1 1 10 n k = a 1 a 2 · · · a n 1 1 1 / 10 n 1 = a 1 a 2 · · · a n 10 n 1 . 12. (a) Denote the partial sums of the first series by s n and those of the second series by t n and observe that s n = ( a 0 + a 1 + · · · + a j ) + t n . Obviously s n L iff t n = s n ( a 0 + a 1 + · · · + a j ) L ( a 0 + a 1 + · · · + a j ) .
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