chapter1Solutions

# chapter1Solutions - Chapter 1 Areas, volumes and simple...

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Unformatted text preview: 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 + 1 = 11 terms. (c) 2 + 2 1 + 2 2 + 2 3 + 2 4 (d) 2 + 2 1 + 2 2 + 2 3 + 2 4 v.2005.1 - December 16, 2008 1 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 parenleftbigg 1 2 parenrightbigg k + 4 summationdisplay k =2 parenleftbigg 1 2 parenrightbigg k (f) Simplify 50 summationdisplay x =0 3 x − 50 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 parenleftbigg 1 2 parenrightbigg k v.2005.1 - December 16, 2008 2 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 - December 16, 2008 3 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...
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## This note was uploaded on 02/09/2011 for the course MATH 102 taught by Professor Lee during the Spring '11 term at Simon Fraser.

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chapter1Solutions - Chapter 1 Areas, volumes and simple...

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