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HW11 - momin(rrm497 Homework 11 cheng(58520 This print-out...

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momin (rrm497) – Homework 11 – cheng – (58520) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite the finite sum 12 3 + 4 16 4 + 4 + 20 5 + 4 24 6 + 4 + . . . + ( 1) 6 36 9 + 4 using summation notation. 1. 10 summationdisplay k =4 ( 1) k 3 4 k k + 4 2. 6 summationdisplay k =0 ( 1) k 3 4 k k + 4 3. 9 summationdisplay k =3 ( 1) k 4 4 k k + 4 4. 9 summationdisplay k =3 ( 1) k 3 4 k k + 4 correct 5. 6 summationdisplay k =0 4 k k + 4 6. 9 summationdisplay k =3 4 k k + 4 Explanation: The numerators form a sequence 12 , 16 , 20 , 24 , . . . , 36 , while the denominators form a sequence 3 + 4 , 4 + 4 , 5 + 4 , 6 + 4 , . . . , 9 + 4 . Thus the general term in the series is of the form a k = ( 1) k 3 4 k k + 4 where the sum ranges from k = 3 to k = 9. Consequently, the series becomes 9 summationdisplay k =3 ( 1) k 3 4 k k + 4 in summation notation. 002 10.0 points If the n th partial sum of n =1 a n is given by S n = 4 n + 1 n + 4 , what is a n when n 2? 1. a n = 15 n ( n + 4) 2. a n = 17 ( n + 4)( n + 3) 3. a n = 15 ( n + 4)( n + 5) 4. a n = 17 n ( n + 4) 5. a n = 17 ( n + 4)( n + 5) 6. a n = 15 ( n + 4)( n + 3) correct Explanation: By definition S n = n summationdisplay k 1 a n = a 1 + a 2 + . . . + a n . Thus, for n 2, a n = S n S n 1 = 4 n + 1 n + 4 4( n 1) + 1 ( n 1) + 4 . Consequently, a n = 15 ( n + 4)( n + 3) . 003 10.0 points
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momin (rrm497) – Homework 11 – cheng – (58520) 2 Determine whether the series summationdisplay n =0 3 (cos ) parenleftbigg 1 3 parenrightbigg n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum 4 9 2. convergent with sum 9 2 3. convergent with sum 9 4 4. divergent 5. convergent with sum 9 2 6. convergent with sum 9 4 correct Explanation: Since cos = ( 1) n , the given series can be rewritten as an infinite geometric series summationdisplay n =0 3 parenleftbigg 1 3 parenrightbigg n = summationdisplay n =0 a r n in which a = 3 , r = 1 3 . But the series n =0 ar n is (i) convergent with sum a 1 r when | r | < 1, and (ii) divergent when | r | ≥ 1. Consequently, the given series is convergent with sum 9 4 . 004 10.0 points Determine whether the infinite series summationdisplay n =1 3( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum. 1. converges with sum = 3 2. converges with sum = 3 4 3. diverges correct 4. converges with sum = 3 8 5. converges with sum = 3 2 Explanation: By the Divergence Test, an infinite series n a n diverges when lim n →∞ a n negationslash = 0 . Now, for the given series, a n = 3( n + 1) 2 n ( n + 2) = 3 n 2 + 6 n + 3 n 2 + 2 n . But then, lim n → ∞ a n = 3 negationslash = 0 . Consequently, the Divergence Test says that the given series diverges . keywords: infinite series, Divergence Test, ra- tional function 005 10.0 points Determine whether the infinite series summationdisplay n =1 tan 1 parenleftBig 2 n 2 25 n + 1 parenrightBig
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momin (rrm497) – Homework 11 – cheng – (58520) 3 converges or diverges, and if it converges, find its sum.
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