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

This preview shows pages 1–4. Sign up to view the full content.

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 ,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern