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int cal 11pdf

# int cal 11pdf - pokharel(yp624 HW11 Radin(57410 This...

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pokharel (yp624) – HW11 – Radin – (57410) 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 Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 4 2 n + 4 parenrightbigg , and if it does, find its limit. 1. the sequence diverges 2. limit = ln 2 3. limit = ln 2 3 4. limit = ln 2 5. limit = 0 correct Explanation: After division by n we see that 4 2 n + 4 = 4 n 2 + 4 n , so by properties of logs, a n = 1 n ln 4 n 1 n ln parenleftbigg 2 + 4 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 4 n , 1 n ln parenleftbigg 2 + 4 n parenrightbigg −→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 2 , 6 , 10 , 14 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = n + 4 2. a n = n + 3 3. a n = 4 n 2 correct 4. a n = 5 n 3 5. a n = 3 n 1 Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 2 , 6 , 10 , 14 , . . . 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 = 2 + 4( n 1) . Consequently, a n = 4 n 2 . keywords: 003 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 5 3 , 25 9 , 125 27 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 5 3 parenrightBig n 2. a n = parenleftBig 3 5 parenrightBig n 1

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pokharel (yp624) – HW11 – Radin – (57410) 2 3. a n = parenleftBig 3 2 parenrightBig n 1 4. a n = parenleftBig 3 5 parenrightBig n 5. a n = parenleftBig 3 2 parenrightBig n 6. a n = parenleftBig 5 3 parenrightBig n 1 correct Explanation: By inspection, consecutive terms a n 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 5 3 , 25 9 , 125 27 , . . . bracerightBig have the property that a n = ra n 1 = parenleftBig 5 3 parenrightBig a n 1 . Thus a n = ra n 1 = r 2 a n 2 = . . . = r n 1 a 1 = parenleftBig 5 3 parenrightBig n 1 a 1 . Consequently, a n = parenleftBig 5 3 parenrightBig n 1 since a 1 = 1. keywords: sequence, common ratio 004 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 3 n 5 n 3 + 4 7 n 4 + 4 n 2 + 2 . 1. limit = 1 4 2. limit = 2 3. the sequence diverges correct 4. limit = 0 5. limit = 3 7 Explanation: After division by n 4 we see that a n = 3 n 1 n + 4 n 4 7 + 4 n 2 + 2 n 4 . Now 1 n , 4 n 4 , 4 n 2 , 2 n 4 −→ 0 as n → ∞ ; in particular, the denominator converges and has limit 7 negationslash = 0. Thus by properties of limits { a n } diverges since the sequence { 3 n } diverges.
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int cal 11pdf - pokharel(yp624 HW11 Radin(57410 This...

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