HW 2-solutions

# HW 2-solutions - morris(gmm643 – HW 2 – mann –(54675...

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Unformatted text preview: morris (gmm643) – HW 2 – mann – (54675) 1 This print-out should have 18 questions. Multiple-choice 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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HW 2-solutions - morris(gmm643 – HW 2 – mann –(54675...

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