Chapter 5 - SECTION 5.1(PAGE 278 R A ADAMS CALCULUS CHAPTER...

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SECTION 5.1 (PAGE 278) R. A. ADAMS: CALCULUS CHAPTER 5. INTEGRATION Section 5.1 Sums and Sigma Notation (page 278) 1. 4 ± i = 1 i 3 = 1 3 + 2 3 + 3 3 + 4 3 2. 100 ± j = 1 j j + 1 = 1 2 + 2 3 + 3 4 +···+ 100 101 3. n ± i = 1 3 i = 3 + 3 2 + 3 3 3 n 4. n 1 ± i = 0 ( 1 ) i i + 1 = 1 1 2 + 1 3 −···+ ( 1 ) n 1 n 5. n ± j = 3 ( 2 ) j ( j 2 ) 2 =− 2 3 1 2 + 2 4 2 2 2 5 3 2 ( 1 ) n 2 n ( n 2 ) 2 6. n ± j = 1 j 2 n 3 = 1 n 3 + 4 n 3 + 9 n 3 n 2 n 3 7. 5 + 6 + 7 + 8 + 9 = 9 ± i = 5 i 8. 2 + 2 + 2 2 (200 terms) equals 200 ± i = 1 2 9. 2 2 3 2 + 4 2 5 2 +···− 99 2 = 99 ± i = 2 ( 1 ) i i 2 10. 1 + 2 x + 3 x 2 + 4 x 3 100 x 99 = 100 ± i = 1 ix i 1 11. 1 + x + x 2 + x 3 x n = n ± i = 0 x i 12. 1 x + x 2 x 3 x 2 n = 2 n ± i = 0 ( 1 ) i x i 13. 1 1 4 + 1 9 ( 1 ) n 1 n 2 = n ± i = 1 ( 1 ) i 1 i 2 14. 1 2 + 2 4 + 3 8 + 4 16 n 2 n = n ± i = 1 i 2 i 15. 99 ± j = 0 sin j = 100 ± i = 1 sin ( i 1 ) 16. m ± k =− 5 1 k 2 + 1 = m + 6 ± i = 1 1 (( i 6 ) 2 + 1 17. n ± i = 1 ( i 2 + 2 i ) = n ( n + 1 )( 2 n + 1 ) 6 + 2 n ( n + 1 ) 2 = n ( n + 1 )( 2 n + 7 ) 6 18. 1 , 000 ± j = 1 ( 2 j + 3 ) = 2 ( 1 , 000 )( 1 , 001 ) 2 + 3 , 000 = 1 , 004 , 000 19. n ± k = 1 k 3 ) = π(π n 1 ) π 1 3 n 20. n ± i = 1 ( 2 i i 2 ) = 2 n + 1 2 1 6 n ( n + 1 )( 2 n + 1 ) 21. n ± m = 1 ln m = ln 1 + ln 2 ln n = ln ( n ! ) 22. n ± i = 0 e i / n = e ( n + 1 )/ n 1 e 1 / n 1 23. 2 + 2 2 (200 terms) equals 400 24. 1 + x + x 2 x n = ² 1 x n + 1 1 x if x ±= 1 n + 1i f x = 1 25. 1 x + x 2 x 3 x 2 n = ² 1 + x 2 n + 1 1 + x if x ±=− 1 2 n + f x 1 26. Let f ( x ) = 1 + x + x 2 x 100 = x 101 1 x 1 if x ±= 1. Then f ² ( x ) = 1 + 2 x + 3 x 2 100 x 99 = d dx x 101 1 x 1 = 100 x 101 101 x 100 + 1 ( x 1 ) 2 . 27. 2 2 3 2 + 4 2 5 2 98 2 99 2 = 49 ± k = 1 [ ( 2 k ) 2 ( 2 k + 1 ) 2 ] = 49 ± k = 1 [4 k 2 4 k 2 4 k 1] 49 ± k = 1 [4 k + 1] 4 49 × 50 2 49 4 , 949 28. Let s = 1 2 + 2 4 + 3 8 n 2 n . Then s 2 = 1 4 + 2 8 + 3 16 n 2 n + 1 . 176
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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.1 (PAGE 278) Subtracting these two sums, we get s 2 = 1 2 + 1 4 + 1 8 +···+ 1 2 n n 2 n + 1 = 1 2 1 ( 1 / 2 n ) 1 ( 1 / 2 ) n 2 n + 1 = 1 n + 2 2 n + 1 . Thus s = 2 + ( n + 2 )/ 2 n . 29. n ± i = m ² f ( i + 1 ) f ( i ) ³ = n ± i = m f ( i + 1 ) n ± i = m f ( i ) = n + 1 ± j = m + 1 f ( j ) n ± i = m f ( i ) = f ( n + 1 ) f ( m ), because each sum has only one term that is not cancelled by a term in the other sum. It is called “telescoping” because the sum “folds up” to a sum involving only part of the first and last terms. 30. 10 ± n = 1 ( n 4 ( n 1 ) 4 = 10 4 0 4 = 10 , 000 31. m ± j = 1 ( 2 j 2 j 1 ) = 2 m 2 0 = 2 m 1 32. 2 m ± i = m ´ 1 i 1 i + 1 µ = 1 m 1 2 m + 1 = m + 1 m ( 2 m + 1 ) 33. m ± j = 1 1 j ( j + 1 ) = m ± j = 1 ´ 1 j 1 j + 1 µ = 1 1 n + 1 = n n + 1 34. The number of small shaded squares is 1 + 2 n . Since each has area 1, the total area shaded is n i = 1 i . But this area consists of a large right-angled triangle of area n 2 / 2 (below the diagonal), and n small triangles (above the diagonal) each of area 1/2. Equating these areas, we get n ± i = 1 i = n 2 2 + n 1 2 = n ( n + 1 ) 2 . Fig. 5.1.34 35. To show that n ± i = 1 i = n ( n + 1 ) 2 , we write n copies of the identity ( k + 1 ) 2 k 2 = 2 k + 1 , one for each k from 1 to n : 2 2 1 2 = 2 ( 1 ) + 1 3 2 2 2 = 2 ( 2 ) + 1 4 2 3 2 = 2 ( 3 ) + 1 .
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This note was uploaded on 12/16/2009 for the course FEW, FEWEB 400567 taught by Professor Moerdersen during the Fall '09 term at Vrije Universiteit Amsterdam.

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Chapter 5 - SECTION 5.1(PAGE 278 R A ADAMS CALCULUS CHAPTER...

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