Solutions to Exam #4 for MATH 110 - Section 10, Fall 2005, Exams of Algebra

Download Solutions to Exam #4 for MATH 110 - Section 10, Fall 2005 and more Exams Algebra in PDF only on Docsity! Exam #4 solutions · Fri. Dec. 2, 2005 MATH 110 · Section 10 · Fall 2005 Name REMINDER: The final exam is Monday, December 12, 8:00 a.m. to 10:00 a.m., in CESL 103. Please arrive at 7:40 a.m. Formulas: A = Pert ∑n k=1 k = n(n+1) 2 ∑n k=1 k 2 = n(n+1)(2n+1)6 an = a1 + (n − 1)d Sn = n ( a1+an 2 ) Sn = n2 (2a1 + (n − 1)d) an = a1rn−1 Sn = a1 ( 1−rn 1−r ) , r 6= 1 ∑∞ k=1 a1r k−1 = a11−r , |r| < 1 Problem 1. (6 points) One may approximate the altitude A (in meters above sea level) using the formula A = 22, 860 ln ( p0 p ) where p is the atmospheric pressure in mm Hg at that altitude, and p0 is the sea-level atmospheric pressure, 760 mm Hg. At an altitude of 4,000 meters, what is the approximate atmospheric pressure? Answer. We are given p0 = 760 and A = 4000, and we are asked to find p: 4000 = 22860 ln ( 760 p ) 4000 22860 = ln ( 760 p ) e4000/22860 = 760 p p = 760 e4000/22860 ≈ 638. Problem 2. (6 points) Your aunt Judy gave you a generous high-school graduation gift of $4,000, which you wisely invested in an account yielding 6% annual interest (compounded continuously). How long (in years, to the nearest 0.1 year) will it take for the balance to reach $10,000? Answer. This is a compound-interest problem, so use the compound-interest formula, A = Pert. We are given P = 4000, A = 10000, and r = 0.06, and we are asked to find t: 10000 = 4000e0.06t 10000 4000 = e0.06t 2.5 = e0.06t ln(2.5) = 0.06t t = ln(2.5) 0.06 ≈ 15.3. 2 Problem 3. (6 points) Consider the sequence whose terms given by an = (−1)n n2 + 1 . Which of the following statements are true about the 6th and 7th terms of the sequence? (1)a6 < 0 (2)a7 < 0 (3)a6 < a7 (A) 1, 2, and 3 (B) 1 and 2 only (C) 1 only (D) 2 only (E) None of these Answer. Given the formula an = (−1)n n2 + 1 , we can compute a6 = (−1)6 62 + 1 = 1 37 and a7 = (−1)7 72 + 1 = −1 50 . Since a6 > 0 and a7 < 0, only (2) applies. Problem 4. Consider the sequence 3 2 , 4 4 , 5 6 , 6 8 , 7 10 , . . . . (a) (4 points) Find a closed-form expression for an. (This means that the formula for an should not refer to an−1.) Answer. Write down the table of values n an 1 3/2 2 4/4 3 5/6 4 6/8 5 7/10 The numerators are each 2 more than n, and the denominators are each twice n. So the nth term is an = n + 2 2n . (b) (4 points) What is the 20th term of the sequence? Answer. Evaluate the formula from part (a) at n = 20: a20 = 20 + 2 2 · 20 = 22 40 = 11 20 = 0.55. (c) (4 points) Describe the long-term behavior of the sequence: does it approach a specific value? If so, what and why? If not, why not? Answer. The sequence is the rational function x+22x for positive integer inputs, which has horizontal asymptote 1/2. This can be checked by computing a few more terms of the sequence. 5 Answer. This is an infinite geometric sum with a1 = 4 and r = 1/3, so use the formula for infinite geometric sums from page 1: ∞∑ k=1 4 ( 1 3 )k−1 = ∞∑ k=1 a1r k−1 = a1 1 − r = 4 1 − 13 = 4 2 3 = 4 · 3 2 = 6. Problem 11. (6 points) Find a closed-form expression for the nth term of the arithmetic sequence with first term 7.2 and common difference 0.1. Answer. This is an arithmetic sequence with a1 = 7.2 and d = 0.1. So, use the formula for arithmetic sequences: an = a1 + (n − 1)d = 7.2 + (n − 1)0.1 = 7.2 + 0.1n − 0.1 = 7.1 + 0.1n. Problem 12. (6 points) Find the number of terms in the sequence 11.3, 11.5, 11.7, . . . , 57.5. Answer. This is an arithmetic sequence with a1 = 7.2, an = 0.1, and d = 0.1. So, use the formula for arithmetic sequences: an = a1 + (n − 1)d 57.5 = 11.3 + (n − 1)0.2 46.2 = (n − 1)0.2 46.2 0.2 = n − 1 n − 1 = 231 n = 232. Problem 13. (6 points) Find the sum of the even positive integers up to and including 500. Answer. This is an arithmetic sum with a1 = 2, an = 500, and d = 2. The formula for arithmetic sums requires us to know the number of terms, n. To find out how many terms there are in the sum, use the formula for arithmetic sequences: an = a1 + (n − 1)d 500 = 2 + (n − 1)2 500 = 2 + 2n − 2 500 = 2n n = 250. 6 Then, use the formula for arithmetic sums: Sn = n ( a1 + an 2 ) = 250 ( 2 + 500 2 ) = 250 ( 502 2 ) = 250 · 251 = 62750. Problem 14. An employer in Connecticut has offered you a starting salary of $40,000 per year with 3% annual increases. An employer in San Diego has offered you a starting salary of $35,000 per year with 4% annual increases. (a) (6 points) For the Connecticut job, what would be your total earnings over a 10-year period? Answer. First we need to recognize what kind of problem this is. • The salary in year 1 is 40, 000. • The salary in year 2 is 40, 000 · 1.03. • The salary in year 3 is (40, 000 · 1.03) · 1.03 = 40, 000 · 1.032. • Likewise, the salary in year n is 40000 · 1.03n−1. Since each year’s salary is 1.03 times the previous year’s salary, and since the question is to find the total earnings over a 10-year period, which is the sum of all the annual salaries, we have a geometric sum. Use the geometric-sum formula from page 1: Sn = a1 ( 1 − rn 1 − r ) S10 = a1 ( 1 − r10 1 − r ) = 40000 ( 1 − 1.0310 1 − 1.03 ) ≈ 458, 555. (b) (6 points) For the San Diego job, what would be your total earnings over a 10-year period? Answer. Similar to the first part, Sn = a1 ( 1 − rn 1 − r ) S10 = a1 ( 1 − r10 1 − r ) = 35000 ( 1 − 1.0410 1 − 1.04 ) ≈ 420, 213. Problem 15. (6 points) As you rise through management, you will need to deal with other people’s annoying problems more and more often. If you deal with 8 annoyances per month in your entry-level position, and if the number of annoyances per month increases by 20 percent for each level you rise in the company, how many annoyances per month can you expect once you reach regional vice president, which is the seventh layer of management? Answer. Again, first there is the recognition problem: what kind of problem is this? • At the first level of the company, you have 8 annoynances per month. 7 • At the second level of the company, you have 8 · 1.20 annoynances per month. • At the third level of the company, you have (8 · 1.20) · 1.2 = 8 · 1.202 annoynances per month. • ... • At the seventh level of the company, you have 8 · 1.206 annoynances per month. This is a sequence of numbers, with each one 1.20 times the previous. That means it’s a geometric sequence, and we just need to find the seventh term. Use the formula for geometric sequences: an = a1rn−1 a7 = a1r7−1 = 8 · 1.206 ≈ 24.