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Unformatted text preview: morris (gmm643) HW 2 mann (54675) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Sections 12.2, 12.3 001 10.0 points Rewrite the finite sum 4 2 + 1 + 6 3 + 1 + 8 4 + 1 + 10 5 + 1 + . . . + 16 8 + 1 using summation notation. 1. 9 summationdisplay k = 3 2 k k + 1 2. 6 summationdisplay k = 0 2 k k + 1 3. 8 summationdisplay k = 2 2 k k + 1 correct 4. 9 summationdisplay k = 3 ( 1) k 2 2 k k + 1 5. 8 summationdisplay k = 2 ( 1) k 2 2 k k + 1 6. 6 summationdisplay k = 0 ( 1) k 2 2 k k + 1 Explanation: The numerators form a sequence 4 , 6 , 8 , 10 , . . . , 16 , while the denominators form a sequence 2 + 1 , 3 + 1 , 4 + 1 , 5 + 1 , . . . , 8 + 1 . Thus the general term in the series is of the form a k = 2 k k + 1 where the sum ranges from k = 2 to k = 8. Consequently, the series becomes 8 summationdisplay k =2 2 k k + 1 in summation notation. 002 (part 1 of 3) 10.0 points Write each of the following finite sums in summation notation. (i) The sum of the first ten positive odd inte gers. 1. sum = 10 summationdisplay i =1 (2 i 1) correct 2. sum = 10 summationdisplay i =1 (2 i + 1) 3. sum = 10 summationdisplay i =1 i 4. sum = 10 summationdisplay i =1 2 i 5. sum = 10 summationdisplay i =1 ( i 1) Explanation: Positive odd integers are represented by 2 i 1, i = 1 , 2 , 3 , . . . . Thus the required sum is given by sum = 10 summationdisplay i =1 (2 i 1). 003 (part 2 of 3) 10.0 points (ii) The sum of the cubes of the first n positive integers. 1. sum = n summationdisplay i =1 i 2. None of these morris (gmm643) HW 2 mann (54675) 2 3. sum = n summationdisplay i =0 i 3 4. sum = n summationdisplay i =0 i 5. sum = n summationdisplay i =1 i 3 correct Explanation: The positive integers are represented by i , i = 1 , 2 , 3 , . . . . Thus the required sum is given by sum = n summationdisplay i =1 i 3 . 004 (part 3 of 3) 10.0 points (iii) 6 + 10 + 14 + 18 + . . . + 42 . 1. sum = 10 summationdisplay i =1 6 i 2. sum = 10 summationdisplay i =1 (2 + 4 i ) correct 3. sum = 10 summationdisplay i =1 (6 + 4 i ) 4. sum = 10 summationdisplay i =1 4 i 5. None of these Explanation: The difference between consecutive terms is 4, so 6+10 + 14 + 18 + + 42 = (2 + 4) + (2 + 8) + (2 + 12)+ . . . + (2 + 40) = sum = 10 summationdisplay i =1 (2 + 4 i ). Alternate Solution: 6+10 + 14 + 18 + + 42 = (6 + 0) + (6 + 4) + (6 + 8)+ . . . + (6 + 36) = 10 summationdisplay i =1 [6 + 4 ( i 1)] = sum = 10 summationdisplay i =1 (2 + 4 i ). 005 10.0 points Determine whether the infinite series 3 1 + 1 3 1 9 + 1 27 is convergent or divergent, and if convergent, find its sum....
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 Spring '10
 COKER

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