ca1-2 - Version PREVIEW FinalReview03 Van Ligten(56650 This...

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Version PREVIEW – FinalReview03 – Van Ligten – (56650) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. CalC12b04exam2 001 10.0 points If the n th partial sum of n =1 a n is given by S n = 2 n + 5 n + 4 , what is a n when n 2? 1. a n = 3 ( n + 4)( n + 5) 2. a n = 13 n ( n + 4) 3. a n = 13 ( n + 4)( n + 5) 4. a n = 3 n ( n + 4) 5. a n = 13 ( n + 4)( n + 3) 6. a n = 3 ( n + 4)( n + 3) correct Explanation: By definition S n = n summationdisplay k 1 a n = a 1 + a 2 + . . . + a n . Thus, for n 2, a n = S n S n 1 = 2 n + 5 n + 4 2( n 1) + 5 ( n 1) + 4 . Consequently, a n = 3 ( n + 4)( n + 3) . CalC12c01s 002 10.0 points Compare the values of the series A = summationdisplay n =1 4 n 2 and the improper integral B = integraldisplay 1 4 x 2 dx . 1. A = B 2. A > B correct 3. A < B Explanation: In the figure 1 2 3 4 5 . . . a 1 a 2 a 3 a 4 the bold line is the graph of the function f ( x ) = 4 x 2 on [1 , ) and the area of each of the rectan- gles is one of the values of a n = 4 n 2 . Clearly from this figure we see that a 1 = f (1) > integraldisplay 2 1 f ( x ) dx, a 2 = f (2) > integraldisplay 3 2 f ( x ) dx ,
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Version PREVIEW – FinalReview03 – Van Ligten – (56650) 2 while a 3 = f (3) > integraldisplay 4 3 f ( x ) dx, a 4 = f (4) > integraldisplay 5 4 f ( x ) dx , and so on for all n . Consequently, A = summationdisplay n =1 4 n 2 > integraldisplay 1 4 x 2 dx = B . CalC12b44b 003 10.0 points Find the sum of the series summationdisplay n =0 (2 x 5) n 4 n for those values of x for which it converges. 1. sum = 1 + 2 x 4 2. sum = 4 1 2 x 3. sum = 4 9 2 x correct 4. sum = 4 9 + 2 x 5. sum = 4 1 + 2 x 6. sum = 9 2 x 4 Explanation: When the geometric series n =0 ar n con- verges it has sum = a 1 r . Now for the given series, a = 1 , r = 2 x 5 4 , Consequently, it has sum = 1 1 2 x 5 4 = 4 9 2 x . CalC12c11s 004 10.0 points Determine the convergence or divergence of the series ( A ) 1 + 1 4 + 1 9 + 1 16 + 1 25 + ... , and ( B ) summationdisplay n =1 n 2 e n 3 . 1. A divergent, B convergent 2. A convergent, B divergent 3. both series convergent correct 4. both series divergent Explanation: ( A ) The given series has the form 1 + 1 4 + 1 9 + 1 16 + 1 25 + ... = summationdisplay n =1 1 n 2 . This is a p -series with p = 2 > 1, so the series converges. ( B ) The given series has the form summationdisplay n =1 f ( n ) with f defined by f ( x ) = x 2 e x 3 . Note first that f is continuous and positive on [1 , ); in addition, since f ( x ) = e x 3 (2 x 1 3 x 4 ) < 0
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Version PREVIEW – FinalReview03 – Van Ligten – (56650) 3 for x > 1, f is decreasing on [1 , ). Thus we can use the Integral Test. Now, by substitu- tion, integraldisplay t 1 x 2 e x 3 dx = bracketleftbigg 1 3 e x 3 bracketrightbigg t 1 , and so integraldisplay 1 x 2 e x 3 dx = 1 3 e . Since the integral converges, the series con- verges. This could also be established using the Ratio Test.
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