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Unformatted text preview: Valencia (drv252) Review 3 Zheng (58355) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the n th term, a n , of an infinite series n = 1 a n when the n th partial sum, S n , of the series is given by S n = 4 n n + 1 . 1. a n = 5 n ( n + 1) 2. a n = 5 2 n 3. a n = 2 n 4. a n = 4 n ( n + 1) correct 5. a n = 5 2 n 2 6. a n = 2 n 2 Explanation: Since S n = a 1 + a 2 + + a n , we see that a 1 = S 1 , a n = S n S n 1 ( n > 1) . But S n = 4 n n + 1 = 4 4 n + 1 . Thus a 1 = 2, while a n = 4 n 4 n + 1 , ( n > 1) . Consequently, a n = 4 n 4 n + 1 = 4 n ( n + 1) for all n . 002 10.0 points Let f be a continuous, positive, decreasing function on [3 , ). Compare the values of the integral A = integraldisplay 20 3 f ( t ) dt and the series B = 19 summationdisplay n = 3 f ( n ) . 1. A = B 2. A > B 3. A < B correct Explanation: In the figure 3 4 5 6 7 . . . a 3 a 4 a 5 a 6 the bold line is the graph of f on [3 , ) and the areas of the rectangles the terms in the series summationdisplay n = 3 a n , a n = f ( n ) . Clearly from this figure we see that f (3) > integraldisplay 4 3 f ( t ) dt, f (4) > integraldisplay 5 4 f ( t ) dt , Valencia (drv252) Review 3 Zheng (58355) 2 while f (5) > integraldisplay 6 5 f ( t ) dt, f (6) > integraldisplay 7 6 f ( t ) dt , and so on. Consequently, A < B . keywords: 003 10.0 points To apply the root test to an infinite series k a k , the value of = lim k  a k  1 /k has to be determined. Compute the value of for the series summationdisplay k = 1 3 k k (ln k + 7) k . 1. = 7 2. = 21 3. = 3 4. = 5. = 0 correct Explanation: For the given series ( a k ) 1 /k = 3 1 /k parenleftbigg ln k + 7 k parenrightbigg = 3 1 /k parenleftbigg ln k k + 7 k parenrightbigg . But 3 1 /k 1 , ln k k as k . Consequently, = 0 . 004 10.0 points Determine whether the series summationdisplay n =0 parenleftbigg 4 7 parenrightbigg n/ 2 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 7 2 7 2. convergent with sum = 7 2 2 3. convergent with sum = 7 7 2 correct 4. convergent with sum = 2 7 2 5. divergent Explanation: The infinite series summationdisplay n =0 parenleftbigg 4 7 parenrightbigg n/ 2 is an infinite geometric series n =0 ar n with a = 1 and r = 2 / 7. But n =0 ar n is (i) convergent with sum a 1 r when  r  < 1, and (ii) divergent when  r  1 . So the given series is convergent with sum = 7 7 2 ....
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 Spring '08
 Cepparo

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