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Unformatted text preview: husain (aih243) 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 12 3 + 1 + 16 4 + 1 + 20 5 + 1 + 24 6 + 1 + . . . + 32 8 + 1 using summation notation. 1. 8 summationdisplay k = 3 4 k k + 1 correct 2. 5 summationdisplay k = 0 ( 1) k 3 4 k k + 1 3. 8 summationdisplay k = 3 ( 1) k 3 4 k k + 1 4. 9 summationdisplay k = 4 4 k k + 1 5. 5 summationdisplay k = 0 4 k k + 1 6. 9 summationdisplay k = 4 ( 1) k 3 4 k k + 1 Explanation: The numerators form a sequence 12 , 16 , 20 , 24 , . . . , 32 , while the denominators form a sequence 3 + 1 , 4 + 1 , 5 + 1 , 6 + 1 , . . . , 8 + 1 . Thus the general term in the series is of the form a k = 4 k k + 1 where the sum ranges from k = 3 to k = 8. Consequently, the series becomes 8 summationdisplay k =3 4 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 2. sum = 10 summationdisplay i =1 ( i 1) 3. sum = 10 summationdisplay i =1 (2 i 1) correct 4. sum = 10 summationdisplay i =1 (2 i + 1) 5. sum = 10 summationdisplay i =1 i 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 =0 i 2. sum = n summationdisplay i =1 i 3 correct husain (aih243) HW 2 mann (54675) 2 3. None of these 4. sum = n summationdisplay i =0 i 3 5. sum = n summationdisplay i =1 i 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 4 i 3. sum = 10 summationdisplay i =1 (6 + 4 i ) 4. None of these 5. sum = 10 summationdisplay i =1 (2 + 4 i ) correct 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 2 + 4 3 8 9 + 16 27 is convergent or divergent, and if convergent, find its sum....
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This note was uploaded on 11/30/2010 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler

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