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chapter1ProblemsAndSolutions

chapter1ProblemsAndSolutions - Chapter 1 Areas volumes and...

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Chapter 1 Areas, volumes and simple sums 1.1 Answer the following questions: (a) What is the value of the fifth term of the sum S = 20 k =1 (5 + 3 k ) /k ? (b) How many terms are there in total in the sum S = 17 k =7 e k ? (c) Write out the terms in 5 n =1 2 n - 1 . (d) Write out the terms in 4 n =0 2 n . (e) Write the series 1 + 3 + 3 2 + 3 3 in summation notation in two equivalent forms. Detailed Solution: (a) a 5 = (5 + 3 × 5) / 5 = 4. (b) First term has index k = 7, and last term k = 17. Thus there are 17 - 7 = 10 terms. (c) 2 0 + 2 1 + 2 2 + 2 3 + 2 4 (d) 2 0 + 2 1 + 2 2 + 2 3 + 2 4 v.2005.1 - January 3, 2006 1
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Math 103 Problems Chapter 1 (e) 3 n =0 3 n and 4 n =1 3 n - 1 . 1.2 Summation notation (a) Write 2 + 4 + 6 + 8 + 10 + 12 + ... in summation notation. (b) Write 1 + 1 2 + 1 3 + 1 4 + ... in summation notation. (c) Write out the first few terms of 100 summationdisplay i =0 3 i (d) Write out the first few terms of summationdisplay n =1 1 n n (e) Simplify summationdisplay k =5 2 k + 4 summationdisplay k =2 2 k (f) Simplify summationdisplay x =0 3 x - summationdisplay x =10 3 x (g) Simplify 100 summationdisplay n =0 n + 100 summationdisplay n =0 n 2 (h) Simplify 2 100 summationdisplay y =0 y + 100 summationdisplay y =0 y 2 + 100 summationdisplay y =0 1 Detailed Solution: (a) summationdisplay n =1 2 n (b) summationdisplay n =1 1 n (c) 1 + 3 + 9 + 27 + 81 + 243 + ... (d) 1 + 1 2 2 + 1 3 3 + 1 4 4 + ... (e) summationdisplay k =2 2 k v.2005.1 - January 3, 2006 2
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Math 103 Problems Chapter 1 (f) 9 summationdisplay x =0 3 x (g) 100 summationdisplay n =0 ( n + n 2 ) (h) 100 summationdisplay y =0 ( y + 1) 2 1.3 Show that the following pairs of sequences are equivalent: (a) 10 summationdisplay m =0 ( m + 1) 2 and 11 summationdisplay n =1 n 2 (b) 4 summationdisplay n =1 ( n 2 - 2 n + 1) and 4 summationdisplay n =1 ( n - 1) 2 Detailed Solution: (a) 10 summationdisplay m =0 ( m + 1) 2 = 1 + 2 2 + 3 2 + 4 2 + 5 2 + ... + 11 2 11 summationdisplay n =1 n 2 = 1 + 2 2 + 3 2 + 4 2 + 5 2 + ... + 11 2 (b) 4 summationdisplay n =1 n 2 - 2 n + 1 = (1 - 2 + 1) + (4 - 4 + 1) + (9 - 6 + 1) + (116 - 8 + 1) = 0 + 1 + 4 + 9 4 n =1 ( n - 1) 2 = 0 2 + 1 2 + 2 2 + 3 2 = 0 + 1 + 4 + 9 1.4 Compute the following sums: (a) 290 summationdisplay i =1 1 (b) 150 summationdisplay i =1 2 (c) 80 summationdisplay i =1 3 (d) 50 summationdisplay n =1 n (e) 60 summationdisplay n =1 n (f) 60 summationdisplay n =10 n (g) 100 summationdisplay n =20 n (h) 25 summationdisplay n =1 3 n 2 v.2005.1 - January 3, 2006 3
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Math 103 Problems Chapter 1 (i) 20 summationdisplay n =1 2 n 2 (j) 55 summationdisplay i =1 ( i + 2) (k) 75 summationdisplay i =1 ( i + 1) (l) 500 summationdisplay k =100 k (m) 100 summationdisplay k =50 k (n) 50 summationdisplay k =2 ( k 2 - 2 k + 1) (o) 50 summationdisplay k =5 ( k 2 - 2 k + 1) (p) 20 summationdisplay m =10 m 3 (q) 15 summationdisplay m =0 ( m + 1) 3 For the solutions to these, we will use several summation formulae, and the notation shown below for convenience: S 0 ( n ) = n summationdisplay i =1 1 = n S 1 ( n ) = n summationdisplay i =1 i = n ( n + 1) 2 S 2 ( n ) = n summationdisplay i =1 i 2 = n ( n + 1)(2 n + 1) 6 S 3 ( n ) = n summationdisplay i =1 i 3 = bracketleftbigg n ( n + 1) 2 bracketrightbigg 2 Detailed Solution: (a) 290 i =1 1 = 290 (b) 150 i =1 2 = 2 150 i =1 1 = 2(150) = 300 (c) 80 i =1 3 = 3(80) = 240 (d) 50 n =1 n = (50)(51) / 2 = 1275 (e) 60 n =1 n = 60(61) / 2 = 1830 (f) 60 n =10 n = 60 n =1 n - 9 n =1 n = (60)(61) / 2 - (9)(10) / 2 = 1830 - 45 = 1785 (g) 100 n =20 n = 100 n =1 n - 19 n =1 n = (100)(101) / 2 - (19)(20) / 2 = 5050 - 190 = 4860 (h) 25 n =1 3 n 2 = 3 25 n =1 n 2 = 3 (25)(25+1)(2 × 25+1) 6 = 16575 (i) 20 n =1 2 n
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