HW11-solutions - kim (tk5895) HW11 Henry (54974) 1 This...

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Unformatted text preview: kim (tk5895) HW11 Henry (54974) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 5 , 9 , 13 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = 4 n 3 correct 2. a n = 5 n 4 3. a n = n + 4 4. a n = n + 3 5. a n = 3 n 2 Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 5 , 9 , 13 , . . . bracerightBig have the property that a n a n 1 = d = 4 . Thus a n = a n 1 + d = a n 2 + 2 d = . . . = a 1 + ( n 1) d = 1 + 4( n 1) . Consequently, a n = 4 n 3 . keywords: 002 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 3 5 , 9 25 , 27 125 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 2 3 parenrightBig n 1 2. a n = parenleftBig 5 3 parenrightBig n 3. a n = parenleftBig 5 3 parenrightBig n 1 4. a n = parenleftBig 3 5 parenrightBig n 1 correct 5. a n = parenleftBig 2 3 parenrightBig n 6. a n = parenleftBig 3 5 parenrightBig n Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 3 5 , 9 25 , 27 125 , . . . bracerightBig have the property that a n = ra n 1 = parenleftBig 3 5 parenrightBig a n 1 . Thus a n = ra n 1 = r 2 a n 2 = . . . = r n 1 a 1 = parenleftBig 3 5 parenrightBig n 1 a 1 . Consequently, a n = parenleftBig 3 5 parenrightBig n 1 since a 1 = 1. keywords: sequence, common ratio 003 10.0 points kim (tk5895) HW11 Henry (54974) 2 Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 2 4 n + 1 parenrightbigg , and if it does, find its limit. 1. the sequence diverges 2. limit = ln 4 3. limit = ln 2 5 4. limit = ln 1 2 5. limit = 0 correct Explanation: After division by n we see that 2 4 n + 1 = 2 n 4 + 1 n , so by properties of logs, a n = 1 n ln 2 n 1 n ln parenleftbigg 4 + 1 n parenrightbigg . But by known limits (or use LHospital), 1 n ln 2 n , 1 n ln parenleftbigg 4 + 1 n parenrightbigg as n . Consequently, the sequence { a n } converges and has limit = 0 . 004 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 7 n 5 3 n 3 + 5 6 n 4 + n 2 + 5 . 1. limit = 3 2. limit = 7 6 3. limit = 1 4. the sequence diverges correct 5. limit = 0 Explanation: After division by n 4 we see that a n = 7 n 3 n + 5 n 4 6 + 1 n 2 + 5 n 4 ....
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This note was uploaded on 04/09/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas at Austin.

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HW11-solutions - kim (tk5895) HW11 Henry (54974) 1 This...

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