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HW11-solutions

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adams (bda255) – HW11 – Radin – (56635) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Compute the value of lim n →∞ 3 a n b n 5 a n 2 b n when lim n →∞ a n = 6 , lim n →∞ b n = 2 . 1. limit = 19 17 2. limit = 18 17 3. limit doesn’t exist 4. limit = 19 17 5. limit = 18 17 correct Explanation: By properties of limits lim n 2 3 a n b n = 3 lim n →∞ a n lim n →∞ b n = 36 while lim n →∞ (5 a n 2 b n ) = 5 lim n →∞ a n 2 lim n →∞ b n = 34 negationslash = 0 . Thus, by properties of limits again, lim n →∞ 3 a n b n 5 a n 2 b n = 18 17 . 002 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 2 3 , 4 9 , 8 27 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 3 4 parenrightBig n 2. a n = parenleftBig 2 3 parenrightBig n 3. a n = parenleftBig 2 3 parenrightBig n - 1 correct 4. a n = parenleftBig 3 2 parenrightBig n - 1 5. a n = parenleftBig 3 4 parenrightBig n - 1 6. a n = parenleftBig 3 2 parenrightBig n Explanation: By inspection, consecutive terms a n - 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 2 3 , 4 9 , 8 27 , . . . bracerightBig have the property that a n = ra n - 1 = parenleftBig 2 3 parenrightBig a n - 1 . Thus a n = ra n - 1 = r 2 a n - 2 = . . . = r n - 1 a 1 = parenleftBig 2 3 parenrightBig n - 1 a 1 . Consequently, a n = parenleftBig 2 3 parenrightBig n - 1 since a 1 = 1. keywords: sequence, common ratio 003 10.0 points

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adams (bda255) – HW11 – Radin – (56635) 2 Determine whether the sequence { a n } con- verges or diverges when a n = 6 n 2 6 n + 3 n 2 + 4 n + 1 , and if it does, find its limit 1. the sequence diverges 2. limit = 1 2 correct 3. limit = 1 4 4. limit = 0 5. limit = 1 6 Explanation: After bringing the two terms to a common denominator we see that a n = 6 n 3 + 6 n 2 (6 n + 3) ( n 2 + 4 ) (6 n + 3) ( n + 1) = 3 n 2 24 n 12 6 n 2 + 9 n + 3 . Thus a n = 3 24 n 12 n 2 6 + 9 n + 3 n 2 . But 24 n , 12 n 2 , 9 n , 3 n 2 −→ 0 as n → ∞ . Thus { a n } converges and has limit = 1 2 . 004 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 6 n + ( 1) n 2 n + 1 .
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