HW 11 key

# HW 11 key - Miller Kierste Homework 11 Due 3:00 am Inst...

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

Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Rewrite the finite sum 15 3 + 3 - 20 4 + 3 + 25 5 + 3 - 30 6 + 3 + . . . + ( - 1) 5 40 8 + 3 using summation notation. 1. 5 X k = 0 ( - 1) k - 3 5 k k + 3 2. 8 X k = 3 5 k k + 3 3. 8 X k = 3 ( - 1) k - 4 5 k k + 3 4. 9 X k = 4 ( - 1) k - 3 5 k k + 3 5. 5 X k = 0 5 k k + 3 6. 8 X k = 3 ( - 1) k - 3 5 k k + 3 correct Explanation: The numerators form a sequence 15 , 20 , 25 , 30 , . . . , 40 , while the denominators form a sequence 3 + 3 , 4 + 3 , 5 + 3 , 6 + 3 , . . . , 8 + 3 . Thus the general term in the series is of the form a k = ( - 1) k - 3 5 k k + 3 where the sum ranges from k = 3 to k = 8. Consequently, the series becomes 8 X k = 3 ( - 1) k - 3 5 k k + 3 in summation notation. keywords: finite sum, summation notation, 002 (part 1 of 1) 10 points Find the sum of the finite series 8 + 8 · 5 8 + 8 · 5 2 8 2 + . . . + 8 · 5 8 8 8 . 1. sum = 8 8 8 8 7 - 5 7 3 · 2. sum = 8 8 8 8 8 - 5 8 3 · 3. sum = 8 8 8 8 9 - 5 9 3 · correct 4. sum = 8 8 9 - 5 9 3 · 5. sum = 8 8 8 - 5 8 3 · Explanation: The given series is a finite geometric series 8 X n = 0 ar n , with a = 8 , r = 5 8 . Now 8 X n = 0 ar n = a 1 - r 9 1 - r · . Consequently, sum = 8 8 8 8 9 - 5 9 3 · . keywords: finite geometric series

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

View Full Document
Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 2 003 (part 1 of 3) 10 points If the n th partial sum of n = 1 a n is S n = 2 n - 5 n + 1 , (i) what is a 1 ? 1. a 1 = 7 2 2. a 1 = - 3 2 correct 3. a 1 = 3 2 4. a 1 = 2 5. a 1 = - 7 2 Explanation: Since a 1 = S 1 , a 1 = - 3 2 . 004 (part 2 of 3) 10 points (ii) What is a n for n > 1? 1. a n = 3 n 2 2. a n = 7 n 2 3. a n = 3 n ( n - 1) 4. a n = 7 n ( n + 1) correct 5. a n = 7 n ( n - 1) 6. a n = 3 n ( n + 1) Explanation: Since S n = a 1 + a 2 + · · · + a n , we see that a n = S n - S n - 1 . But S n = 2 n - 5 n + 1 = 2( n + 1) - 7 n + 1 = 2 - 7 n + 1 . Consequently, a n = 7 n - 7 n + 1 = 7 n ( n + 1) for all n > 1. 005 (part 3 of 3) 10 points (iii) What is the sum n = 1 a n ? 1. sum = - 1 2. sum = 1 3. sum = 2 correct 4. sum = 0 5. sum = - 2 Explanation: By definition sum = lim n →∞ S n = lim n → ∞ 2 n - 5 n + 1 · . Thus sum = 2 . keywords: 006 (part 1 of 1) 10 points Determine whether the series X n = 0 4 (cos ) 1 2 n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum 8
Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 3 2. divergent 3. convergent with sum - 3 8 4. convergent with sum 8 3 correct 5. convergent with sum - 8 3 6. convergent with sum - 8 Explanation: Since cos = ( - 1) n , the given series can be rewritten as an infinite geometric series X n =0 4 - 1 2 n = X n = 0 a r n in which a = 4 , r = - 1 2 . But the series n =0 ar n is (i) convergent with sum a 1 - r when | r | < 1, and (ii) divergent when | r | ≥ 1. Consequently, the given series is convergent with sum 8 3 . keywords: geometric series, convergent 007 (part 1 of 1) 10 points Determine whether the series X n = 1 n 2 5 n 2 + 4 is convergent or divergent, and if convergent, find its sum.

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