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# hw 3 - rhodes(ajr2283 HW03 Gilbert(56380 This print-out...

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rhodes (ajr2283) – HW03 – Gilbert – (56380) 1 This print-out should have 22 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 2 , 6 , 10 , 14 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = 3 n 1 2. a n = n + 3 3. a n = 5 n 3 4. a n = 4 n 2 correct 5. a n = n + 4 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: 002 10.0 points Find a formula for the general term a n of the sequence { a n } n =1 = braceleftBig 1 , 5 2 , 25 4 , 125 8 , . . . bracerightBig , assuming that the pattern of the first few terms continues. 1. a n = parenleftBig 2 5 parenrightBig n - 1 2. a n = parenleftBig 2 5 parenrightBig n 3. a n = parenleftBig 5 2 parenrightBig n - 1 correct 4. a n = parenleftBig 5 2 parenrightBig n 5. a n = parenleftBig 2 parenrightBig n - 1 6. a n = parenleftBig 2 parenrightBig n Explanation: By inspection, consecutive terms a n - 1 and a n in the sequence { a n } n =1 = braceleftBig 1 , 5 2 , 25 4 , 125 8 , . . . bracerightBig have the property that a n = ra n - 1 = parenleftBig 5 2 parenrightBig a n - 1 . Thus a n = ra n - 1 = r 2 a n - 2 = . . . = r n - 1 a 1 = parenleftBig 5 2 parenrightBig n - 1 a 1 . Consequently, a n = parenleftBig 5 2 parenrightBig n - 1 since a 1 = 1. keywords: sequence, common ratio 003 10.0 points

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rhodes (ajr2283) – HW03 – Gilbert – (56380) 2 Compute the value of lim n →∞ a n b n 6 a n 2 b n when lim n →∞ a n = 6 , lim n →∞ b n = 2 . 1. limit doesn’t exist 2. limit = 3 10 3. limit = 7 20 4. limit = 3 10 correct 5. limit = 7 20 Explanation: By properties of limits lim n 2 a n b n = lim n →∞ a n lim n →∞ b n = 12 while lim n →∞ (6 a n 2 b n ) = 6 lim n →∞ a n 2 lim n →∞ b n = 40 negationslash = 0 . Thus, by properties of limits again, lim n →∞ a n b n 6 a n 2 b n = 3 10 . 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 + 3 . 1. converges with limit = 3 correct 2. converges with limit = 5 2 3. sequence does not converge 4. converges with limit = 6 5 5. converges with limit = 7 2 Explanation: After division by n we see that a n = 6 + ( 1) n n 2 + 3 n . But ( 1) n n , 3 n −→ 0 as n → ∞ , so a n 3 as n → ∞ . Conse- quently, the sequence converges and has limit = 3 . 005 10.0 points Determine if the sequence { a n } converges when a n = n 2 n ( n 4) 2 n , and if it does, find its limit 1. sequence diverges 2. limit = e 2 3. limit = e 8 correct 4. limit = e - 2 5. limit = 1 6. limit = e - 8 Explanation:
rhodes (ajr2283) – HW03 – Gilbert – (56380) 3 By the Laws of Exponents, a n = parenleftbigg n 4 n parenrightbigg - 2 n = parenleftbigg 1 4 n parenrightbigg - 2 n = bracketleftBig parenleftBig 1 4 n parenrightBig n bracketrightBig - 2 .

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