# Calculus: One and Several Variables

This preview shows pages 1–4. Sign up to view the full content.

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

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

View Full Document
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
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 ) .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## ch12[1] - P1: PBU/OVY JWDD027-12 P2: PBU/OVY...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online