1. Math 140 Quiz 6 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

2. Problem 1 (12) Find a general term, a n : 4, 16, 64, 256, 1024. Test for arithmetic sequence: n =1234 a n+1 - a n =16-4= 1264-16= 48256-64 =1921024-246 =768 No common difference => not arithmetic sequence Test for geometric sequence: n =1234 a n+1 / a n =16/4 = 464/16 = 4256/64 = 41024/246 = 4 Common ratio r = 4 => a n = a 1 r (n-1) = 4(4) (n-1) = 4 n

3. Problem 2 (19) Find a general term, a n : 0, 2, 6, 12, 20. Approach when choices are given - Test each choice: Choicen =12345OK? A)a n = 4n - 6-2261016NO B)a n = 2n – 202468NO C)a n = n 2 – n0261220YES D)a n = 2 (n-1) –1013715NO

4. Problem 2 cont’d (19) Find a general term, a n : 0, 2, 6, 12, 20. Approach if no choices - Test for arithmetic sequence: n =1234 a n+1 - a n =2 – 0= 26 - 4= 412 - 6= 620 -12= 8 No common difference => not arithmetic sequence Test for geometric sequence: n =1c34 a n+1 / a n = 2/0 =  6/4 = 312/6 = 220/12 = 5/3 No common ratio => not geometric sequence

5. Problem 2 cont’d (19) Find a general term, a n : 0, 2, 6, 12, 20. Note pattern in arithmetic sequence test: a n+1 - a n = 2n. n =1234 a n+1 – a n =2468 2n =2n =2468 Hence, a n = a n -1 + 2( n - 1) = a n -2 + 2( n - 1) + 2( n - 2 ) Thus, a n = a 1 + n ( n - 1) = n 2 - n.

6. Problem 3 (31) Write out the first five terms of the sequence to two decimal places. a n = (-1) (n - 1) (n + 1)/(2n - 1) Put n = 1, 2, 3, 4, 5 in the expression for a n and evaluate to get: a 1 = (-1) (1 - 1) (1 + 1)/(21 - 1) = 2 a 2 = (-1) (2 - 1) (2 + 1)/(22 - 1) = -3/3 = -1 a 3 = (-1) (3 - 1) (3 + 1)/(23 - 1) = 4/5 = 0.80 a 4 = (-1) (4 - 1) (4 + 1)/(24 - 1) = -5/7 = -0.71 a 5 = (-1) (5 - 1) (5 + 1)/(25 - 1) = 6/9 = 0.67

7. Problem 4 (37) Find the first term and give a formula for the sequence: 10th term is 16; 15th term is –29. Note choices are of arithmetic sequence: (a 1 = a) _____________ a n = a + d(n - 1). a 10 = a + d(10 - 1) = 16 (a) a 15 = a + d(15 - 1) = -29 (b) Compute: (b) minus (a) & substitute in (a) => d(14 - 9) = -29 –16 => d = -45/5 = -9 a - 9(9) = 16 => a = 97 a n = a + d(n - 1) = 97 - 9(n - 1) = 106 - 9n

8. Problem 5 (12) Express the sum using summation notation, S = 3 + 12 + 27 +... + 108 Since choices are given, test each choice: ChoiceSOK? A)a k = 3k 2 B)a k = 3k 2 C)a k = k 2 D)a k = 3 2 k NO YES

9. Problem 6 (12) Find the common difference, a n : 3.15, 4.82, 6.49, 8.16,.... n =123 a n+1 – a n =4.82 - 3.15 = 1.67 6.49 - 4.82 = 1.67 8.16 - 6.49 = 1.67 Common difference d = 1.67

10. Problem 7 (31) Find the common ratio for the geometric sequence. If a sequence is not geometric, say so. a n : 1, -3, 9, -27, 81 Test for geometric sequence: n =1234 a n+1 / a n =-3/1= -39/-3= -3-27/9= -381/-27=-3 Common ratio r = -3 => a n = a 1 r (n-1) = (-3) (n-1)

11. Problem 8 (81) Find the requested sum of the arithmetic sequence. Use the formula for the sum of first n integers: Put n = 1368 and find: S 1368 = 1368(1368 + 1)/2 = 936,396

12. Problem 9 (81) Find the sum, if it exists, for the infinite geometric sequence. Factor one power of (1/5) to put into standard infinite geometric series form and use formula for the sum of an infinite geometric series with a = 2/5 & r = 1/5 < 1.

13. Problem 10 (25) Determine whether the given sequence is arithmetic, geometric, or neither. If arithmetic, find the common difference. If geometric, find the common ratio. {a n }= {5n 2 - 4} Because n is squared, {a n } is not an arithmetic sequence. Because n is not as power of a constant, r n, {a n } is not a geometric sequence. Numerical tests for arithmetic & geometric sequences are on next slide.

14. Problem 10 cont’d (25) {a n }= {5n 2 - 4} Arithmetic sequence test: a n+1 - a n =5[(n+1) 2 - n 2 ]= 10n + 5 n =1234 a n+1 - a n =15253545 No common difference => not arithmetic sequence Geometric sequence test: a n+1 /a n =[5(n+1) 2 –4]/(5n 2 -4) n =1234 a n+1 / a n =16/1 = 1641/1676/41121/76 No common ratio => not geometric sequence

15. Problem 11 (44) Write the indicated term of the binomial expansion. (3x + 2) 5 ; 5th term Note: in the formula for the binomial expansion, the m th term has i = m – 1. Thus, the m th term is: Here we have n = 5, m = a = 2, & y = 3x. Thus, Next slide has alternate method.

16. Problem 11 Cont’d (62) Write the indicated term of the binomial expansion. (3x + 2) 5 ; 5th term Here is an alternate method. Consider Pascal’s Triangle where each entry is sum of 2 above. 5 th power => 5 th line Thus, term in (y + a) 5 is 5a 4 y. Here a = 2 & y = 3x => 5a 4 y = 52 4 3x = 240x 5th term => next to last entry

17. Problem 12 (44) Evaluate the expression: P(8, 7). P(n, r) = n!/(n – r)! P(8, 7) = 8!/(8 – 7)! = 87654321!/1! = 40,320

18. Problem 13 (56) Evaluate the expression: P(11, 11). P(n, r) = n!/(n – r)! P(11, 11) = 11!/(11 – 11)! = 11109876543210!/0! = 39,916,800

19. Problem 14 (44) Evaluate the expression: C(9, 3). C(n, r) = n!/[r!(n – r)!] C(9, 3) = 9!/[3!(9 – 3)!] = 9876!/[3216!] = 84

20. Problem 15 (62) In a survey of 48 hospital patients, 15 said they were satisfied with the nursing care, 23 said they were satisfied with the medical treatment, and 5 said they were satisfied with both. a) How many patients were satisfied with neither? b) How many were satisfied with only the medical treatment? A: nursing satisfied Quick solution: B: medical satisfied = 15 + 23 - 5 = 33 48 - 33 = 15 satisfied with neither

21. Universe is 48 people. Satisfied by in A = 15 nursing and in B = 23 medical. In A  B = 5 satisfied by both nursing & medical. In A  B = 15+23-5 = 33 ok with nursing or medical. In (A  B) = 48-33 = 15 satisfied by neither. In B - A  B = 23-5 = 18 only medical satisfied. A= 15B= 23 (A  B)=15 A  B=5 Problem 15 cont’d (62)

22. Problem 16 (69) How many 4-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once. The first code position can be filled in 6 ways, the second in 5, the third in 4, & the fourth in 3. Thus, by the Multiplication Rule there are: 6543 = 360 ways. Or, by the Permutation Rule for 6 objects taken 4 at a time, there are: P(n, r) = n!/(n – r)! = 6!/(6–4)! = 65432! /2! = 6543 = 360 ways.

23. Problem 17 (56) In how many ways can 8 volunteers be assigned to 8 booths for a charity bazaar? The first booth can be staffed in 8 ways, the second in 7, the third in 6,.... By the Multiplication Rule there are: 876...1 = 8! ways. That is, 87654321 = 40,320 ways.

24. Problem 18 (37) Tell whether or not the given model is a probability model. The probabilities in Model I all are in [0, 1] and add to 1. So YES, it is a probability model. The probabilities in Model II all are in [0, 1] but add to 1.27 > 1. So NO, it is not a probability model.

25. Problem 19 (31) A bag contains 7 red marbles, 3 blue marbles, and 1 green marble. What is the probability of choosing a marble that is not blue when one marble is drawn from the bag? There are 7 + 3 + 1 = 11 marbles in the bag. Picking a “not blue” one means picking a red or green one of which there are 8. The probability is, thus, 8/11.

26. Problem 20 (25) Two 6-sided dice are rolled. What is the probability that the sum of the two numbers on the dice will be greater than 10? There are 66 = 36 ways that the dice may fall. To get a sum > 10 one must have a sum of 11 or 12. There are only 3 ways to do this: 5+6, 6+5, or 6+6. The probability is, thus, 3/36 = 1/12.

27. Problem 21 (75) A hamburger shop sells hamburgers with cheese, relish, lettuce, tomato, onion, mustard, or ketchup. How many different hamburgers can be concocted using any 3 of the extras? Since order is unimportant, we seek the number of combinations of 7 dressings taken 3 at a time. By the Combination Rule for 7 objects taken 3 at a time, there are: C(n, r) = n!/[r!(n – r)!] = 7!/[3!(7–3)!] = 7654! /(3214!) = 35 ways.