{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw12 - padilla(tp5647 HW12 Gilbert(56650 This print-out...

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

View Full Document Right Arrow Icon
padilla (tp5647) – HW12 – Gilbert – (56650) 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 (part 1 of 3) 10.0 points If the n th partial sum of n =1 a n is S n = 2 n 3 n + 1 , (i) what is a 1 ? 1. a 1 = 1 2 correct 2. a 1 = 2 3. a 1 = 5 2 4. a 1 = 5 2 5. a 1 = 1 2 Explanation: Since a 1 = S 1 , a 1 = 1 2 . 002 (part 2 of 3) 10.0 points (ii) What is a n for n > 1? 1. a n = 1 n 2 2. a n = 5 n ( n + 1) correct 3. a n = 1 n ( n + 1) 4. a n = 1 n ( n 1) 5. a n = 5 n ( n 1) 6. a n = 5 n 2 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 3 n + 1 = 2( n + 1) 5 n + 1 = 2 5 n + 1 . Consequently, a n = 5 n 5 n + 1 = 5 n ( n + 1) for all n > 1. 003 (part 3 of 3) 10.0 points (iii) What is the sum n =1 a n ? 1. sum = 2 correct 2. sum = 1 3. sum = 2 4. sum = 0 5. sum = 1 Explanation: By definition sum = lim n →∞ S n = lim n → ∞ parenleftBig 2 n 3 n + 1 parenrightBig . Thus sum = 2 . 004 10.0 points Determine whether the series summationdisplay n =1 9 n 10 n +1 is convergent or divergent. If it is convergent, find its sum. 1. 65
Background image of page 1

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

View Full Document Right Arrow Icon
padilla (tp5647) – HW12 – Gilbert – (56650) 2 2. 90 3. divergent correct 4. 111 5. 108 Explanation: summationdisplay n =1 9 n 10 n +1 = summationdisplay n =1 10 parenleftbigg 10 9 parenrightbigg n . Since 10 9 > 1, the series is divergent. 005 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 = 9 2 x 4 2. sum = 1 + 2 x 4 3. sum = 4 1 2 x 4. sum = 4 9 2 x correct 5. sum = 4 9 + 2 x 6. sum = 4 1 + 2 x 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 . 006 10.0 points Find the sum of the infinite series tan 2 θ tan 4 θ + tan 6 θ + . . . + ( 1) n 1 tan 2 n θ + . . . whenever the series converges. 1. sum = cos 2 θ 2. sum = sin 2 θ 3. sum = cos 2 θ 4. sum = sin 2 θ correct 5. sum = tan 2 θ Explanation: For general θ the series tan 2 θ tan 4 θ + tan 6 θ + . . . + ( 1) n 1 tan 2 n θ + . . . is an infinite geometric series whose common ratio is tan 2 θ . Since the initial term in this series is tan 2 θ , its sum is thus given by tan 2 θ 1 + tan 2 θ = tan 2 θ sec 2 θ . Consequently sum = sin 2 θ . 007 10.0 points A business executive realizes that he is out of shape so he begins an exercise program in
Background image of page 2
padilla (tp5647) – HW12 – Gilbert – (56650) 3 which he jogs daily and each day thereafter jogs 6% more miles than he did on the previ- ous day. The program will be complete when he has jogged a total of at least 51 miles. If he jogs 2 miles on the first day, what is the mini- mum number of days he will have to exercise to complete the program? 1. 16 Days correct 2. 15 Days 3. 17 Days 4. 18 Days 5. 14 Days Explanation: After the k th day of jogging the executive will have covered a total distance of x = 2 + 2 parenleftBig 1 + 3 50 parenrightBig + 2 parenleftBig 1 + 3 50 parenrightBig 2 + . . . + 2 parenleftBig 1 + 3 50 parenrightBig k 1 , miles in his exercise program. This is a fi-
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}