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 21. Set k = n . Then n = 1 when k = 2 and n = 7 when k =10 . 10 ± k =2 k k 2 = 7 ± n = 1 n ( n 2 = 7 ± n = 1 n n 2 +6 n +10 22. 12 ± n =2 ( 1) n n 1 = 11 ± k =1 ( 1) k +1 k 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 2 3 ³ k = 3 SECTION 12.2 1. 1 2 ; s n = 1 2 ´ 1 1 · 2 + 1 2 · 3 + + 1 ( n )( n +1) µ = 1 2 ´² 1 1 2 ³ + ² 1 2 1 3 ³ + + ² 1 n 1 n +1 ³µ = 1 2 ´ 1 1 n µ 1 2 2. 1 2 ; ± k =3 1 k 2 k = ± k =3 ² 1 k 1 1 k ³ = lim n →∞ ² 1 2 1 n ³ = 1 2 3. 11 18 ; s n = 1 1 · 4 + 1 2 · 5 + + 1 n ( n +3) = 1 3 ´² 1 1 4 ³ + ² 1 2 1 5 ³ + + ² 1 n 1 n +3 ³µ = 1 3 ´ 1+ 1 2 + 1 3 1 n 1 n +2 1 n µ 1 3 ² 1 2 + 1 3 ³ = 11 18 4. 3 4 ; ± k =0 1 ( k + 1)( k = 1 2 ± k =0 ² 1 k 1 k ³ = 1 2 lim n →∞ ² 1 2 1 n 1 n ³ = 3 4 5. 10 3 ; ± k =0 3 10 k =3 ± k =0 ² 1 10 ³ k ² 1 1 1 / 10 ³ = 30 9 = 10 3 6. 5 6 ; ± k =0 ( 1) k 5 k = ± k =0 ² 1 5 ³ k = 1 1 5 = 5 6 7. 3 2 ; ± k =0 1 2 k 3 k = ± k =0 ² 1 3 ³ k ± k =0 ² 2 3 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 ² 3 4 3 ³ k = 3 4 7 · 1 1 3 4 3 = 3 15 , 616 11. Let x =0 . ´ µ¶ · a 1 a 2 ··· a n ´ µ¶ · 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 ² 1 10 n ³ k = a 1 a 2 a n ¸ 1 1 1 / 10 n 1 ¹ = a 1 a 2 a n 10 n 1 . 12. (a) Denote the partial sums of the Frst 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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ch12[1] - P1: PBU/OVY JWDD027-12 P2: PBU/OVY...

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