14_new - HW02 gilbert (55035) This print-out should have 21...

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HW02 – gilbert – (55035) 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 assuming that the pattern of the first few terms continues. 2 n 2. a n = 3 3 n 3. a n = 2 ë o {a n } n=1 = 2, 6, 10, 14, . . . , assuming that the pattern of the first terms continues. 1. a n = n + 3 2. a n = 4n − 2 correct 3. a n = 5n − 3 4. a n = 3n − 1 5. a n = n + 4 Explanation: By inspection, consecutive terms a n−1 a n in the sequence few and 3. a n = 2 3 n−1 4. a n = 3 2 n−1 correct 3 n−1 6. a n = 4 3 n 7. a n = 4 Explanation: By inspection, consecutive terms a n−1 and a n in the sequence {a n } n=1 = ë 1, 2 3 , 4 9 , 27 8 , . . . o ë o {a n } n=1 = 2, 6, 10, 14, . . . have the property that a n − a n−1 = d = 4 . Thus a n = a n−1 + d = a n−2 + 2d = . . . = a 1 + (n − 1)d = 2 + 4(n − 1) . Consequently, a n = 4n − 2 . keywords: 002 10.0 points Find a formula for the general term a n of the sequence {a n } n =1 = ë 1, 3 2 , 9 4 , 27 8 , . . . o , have the property that 2 a n = ra n−1 = 3 a n−1 . Thus a n = ra n−1 = r 2 a n−2 = . . . = r n−1 a 1 = −3 2 n−1 a 1 . Consequently, a n = −3 2 n−1 since a 1 = 1. keywords: sequence, common ratio 003 10.0 points
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2 Compute the value of 5 correct 4a n b n 2. converges with limit = lim 4 when n →∞ 3a n b n 3. converges with limit = 5 7 n lim →∞ a n = 6, n lim →∞ b n = −2. 4. converges with limit = 3 2 1. limit = 12 5 correct 5. sequence does not converge Explanation: 5 After division by n we see that 2. limit = 2 5 + ( −1) n 3. limit doesn’t exist a n = n . 4. limit = 12 4 + n 3 5 But (−1) n , 3 5. limit = 5 −→ 0 2 n n Explanation: as n ∞, so a n 5 4 as n ∞. Conse- quently, the sequence converges and has By properties of limits lim 4a n b n limit = 5 . 4 n→2 = 4 n lim →∞ a n n lim →∞ b n = −48 005 10.0 points while n lim →∞ (3a n b n ) Determine if the sequence {a n } converges when a n = 5 n 2 (2n 1)! , = 3 n lim →∞ a n n lim →∞ b n = 20 6= 0. (2n + 1)! Thus, by properties of limits again, and if it converges, find the limit. 1. converges with limit = 5 correct 4a n b n 12 4 n lim →∞ 3a n − b n = 5 . 2. converges with limit = 2 5 004 10.0 points 5 2 Determine if the sequence {a n } converges, 3. converges with limit = and if it does, find its limit when 4 4. converges with limit = n 5 a n = 5n + (−1) . 4n + 3
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This note was uploaded on 04/04/2012 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.

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14_new - HW02 gilbert (55035) This print-out should have 21...

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