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# Grade 10 Math Module 1 searching for patterns, sequence and series

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### Grade 10 Math Module 1 searching for patterns, sequence and series

1. 1. Module 1 Searching for Patterns in Sequences, Arithmetic, Geometric and Others What this module is all about This module will teach you how to deal with a lot of number patterns. These number patterns are called sequences. Go over the lessons and have fun in working with the exercises. What you are expected to learn It is expected that you will be able to demonstrate knowledge and skill related to sequences and apply these in solving problems. Specifically, you should be able to: a) list the next few terms of a sequence given several consecutive terms. b) derive by pattern searching, a mathematical expression (rule) for generating the sequences. c) generate the next few terms of sequences defined recursively. d) describe an arithmetic sequence by any of the following: • giving the first few terms • giving the formula for the nth term • drawing the graph How much do you know
2. 2. A. Write the first five terms of the sequence. 1. an = n n 3+ 3. an = 4n – 3 2. an = 2n - 1 B. Find the indicated term for the sequence. 4. an = -9n + 2; a8 6. an = 52 73 − + n n ; a14 5. an = (n + 1)(2n + 3); a5 C. Find the general term, an, for the given terms of the sequence. 7. 4, 8, 12, 16,… 9. ,... 12 1 , 6 1 , 2 1 8. -10, -20, -30, -40,… D. Find the first four terms of the sequence defined recursively. 10. a1= -1, an = 3an – 1 11. a1 = 5, an = 1 1 − − n an E. Find the common difference and the next three terms of the given arithmetic sequence. 12. 2, 8, 14, 20,… 14. 1, 2, 3, 4,… 13. 7, 8.5, 10, 11.5,… 15. 24, 21, 18, … What you will do Lesson 1 Sequences and its Kinds It is a common experience to be confronted with a set of numbers arranged in some order. The order and arrangement may be given to you or you have to discover a rule for it from some data. For example, the milkman comes every other day. He came on July 17; will he come on Aug 12? Consider that you are given the set of dates 17, 19, 21,… 2
3. 3. arranged from left to right in the order of increasing time. Continuing the set we have 17, 19, 21, …, 29, 31, 2, 4, ….,28, 30… so that the answer to our question is yes. Any such ordered arrangement of a set of numbers is called a SEQUENCE. Look at this second example. Lorna, a 2nd year student in a certain public school, is able to save the money her ninongs and ninangs gave her last Christmas. She then deposits her savings of P1,000 in an account that earns 10% simple interest. The total amount of interest she earned in each of the 1st 4 years of her saving is shown below: Year 1 2 3 4 Total amount 10 20 30 40 The list of numbers 10, 20, 30, 40 is called a sequence. The list 10,20,30,40 is ordered because the position in this list indicates the year in which that total amount of interest is earned. Now, each of the numbers of a sequence is called a term of the sequence. The first term in the sequence 10, 20, 30, 40 is 10, the second term is 20, while the third term is 30 and the fourth term is 40. It is also good to point out that the preceding term of a given term is the term immediately before that given term. For example, in the given sequence 20 is the term that precedes 30. Examples of other sequences are shown below. These sequences are separated into two groups. A finite sequence contains a finite number of terms. An infinite sequence contains an infinite number of terms. Finite sequence Infinite sequence 1, 1, 2, 3, 5, 8 1, 3, 5, 7, … 1, 2, 3, 4, 5, …, 8 1, , 8 1 , 4 1 , 2 1 … 1, -1, 1, -1 1, 1, 2, 3, 5, 8, … In general: A sequence is a set of numbers written in a specific order: a1, a2, a3, a4, a5, a6,………, an 3
4. 4. The number a1 is called the 1st term, a2 is the 2nd term, and in general, an is the nth term. Note that each term of the sequence is paired with a natural number. Given at least the first 3 terms of a sequence, you can easily find the next term in that sequence by simply discovering a pattern as to how the 3rd term is derived from the 2nd term, and the 2nd from the 1st term. You will find that either a constant number is added or subtracted or multiplied or divided to get the next term or a certain series of operations is performed to get the next term. This may seem hard at first but with practice and patience in getting them, you will find that it’s very exciting. Examples: Find the next term in each sequence. 1. 17, 22, 27, 32, … 2. 11 1 , 8 1 , 5 1 , 2 1 … 3. 5, 10, 20, 40,… 4. 3, -3, 3, -3,… Solutions: 1. Notice that 5 is added to 17 to get 22, the same is added to 22 to get 27, and the same (5) is added to 27 to get 32. So to get the next term add 5 to the preceding term, that is, 32 + 5 = 37. The next term is 37. 2. Notice that 1 is the numerator of all the fractions in the sequence while the denominators- 2, 5, 8, 11 form a sequence. 3 is added to 2 to get 5, 3 is also added to 5 to get 8. So that 3 is added to 11 to get 14. The next term is therefore 1/14. 3. For this example, 2 is multiplied to 5 to get 10, 2 is multiplied to 10 to get 20 and 2 is also multiplied to 20 to get 40. So the next term is 80, the result of multiplying 40 by 2. 4. It is easy to just say that the next term is 3 since the terms in the sequence is alternately positive and negative 3. Actually the first, second, and third terms were multiplied by -1 to get the second, third and fourth terms respectively. Try this out A. Write F if the sequence is finite or I if the sequence is infinite. 1. 2, 3, 4, 5, ….., 10 2. 7, 10, 13, 16, 19, 22, 25 4
5. 5. 3. 4, 9, 14, 19, … 4. 2, 6, 18, 54 5. 3, 9, 27, 81, …., 729, … 6. -2, 4, -8, 16, ….. 7. 100, 97, 94, 91, …, -2 8. 64 1 , 32 1 , 16 1 , 8 1 , 4 1 9. 1, 4, 9, 16, 25, …., 144 10. 25 4 , 16 3 , 9 2 , 4 1 B. Answer the puzzle. Why are Policemen Strong? Find the next number in the sequences and exchange it for the letter which corresponds each sequence with numbers inside the box to decode the answer to the puzzle. A 2, 5, 11, 23, __ N 2, 6, 18, 54, __ B 2, 4, 16, __ O 20, 19, 17, __ C 7, 13, 19, __ P 2, 3, 5, 7, 9, 11, 13, 15, __ D 19, 16, 13, __ R 13, 26, 39, __ E 4, 8, 20, 56, __ S 5, 7, 13, 31, __ F 2, 2, 4, 6, 10, 16, __ T 1, 1, 2, 4, 7, 13, 24, __ H 1, 1, 2, 4, 7, 13, __ U 1, 1, 1, 2, 3, 4, 6, 9, 13, __ I 3, 6, 12, 24, __ Y 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, __ L 10, 11, 9, 12, 8, __ Lesson 2 Finding the Terms of a Sequence Frequently, a sequence has a definite pattern that can be expressed by a rule or formula. In the simple sequence 24 14 13 10 19 17 44 52 47 26 26 48 25 256 164 25 47 19 85 164 44 24 164 6 25 47 162 5
6. 6. 2, 4, 6, 8, 10, …. each term is paired with a natural number by the rule an = 2n. Hence the sequence can be written as 2, 4, 6, 8,… 2n,… 1st term 2nd term 3rd term 4th term nth term a1 a2 a3 a4 an Notice how the formula an = 2n gives all the terms of the sequence. For instance, substituting 1, 2, 3, and 4 for n gives the 1st four terms: a1 = 2(1) = 2 a3 = 2(3) = 6 a2 = 2(2) = 4 a4 = 2(4) = 8 To find the 103rd term of this sequence, use n=103 to get a103 = 2(103) = 206. Examples: 1. Find the 1st four terms of the sequence whose general term is given by an = 2n – 1. Solution: To find the first, second, third and fourth terms of this sequence, simply substitute 1, 2, 3, 4 for n in the formula an = 2n-1. If the general term is an = 2n – 1, then the 1st term is a1 = 2(1) – 1 = 1 2nd term is a2 = 2(2) – 1 = 3 3rd term is a3 = 2(3) – 1 = 5 4th term is a4 = 2(4) – 1 = 7. The 1st four terms of this sequence are the odd numbers 1, 3, 5, and 7. The whole sequence can be written as 1, 3, 5, …, 2n – 1 Since each term in this sequence is larger than the preceding term, we say that the sequence is an increasing sequence. A sequence is increasing if an + 1 > an for all n. 6
7. 7. 2. Write the 1st 4 terms of the sequence defined by an = 1 1 +n . Solution: Replacing n with 1, 2, 3, and 4, respectively the 1st four terms are: 1st term = a1 = 11 1 + = 2 1 2nd term = a2 = 12 1 + = 3 1 3rd term = a3 = 13 1 + = 4 1 4th term = a4 = 14 1 + = 5 1 The sequence defined by an = 1 1 +n can be written as 2 1 , 3 1 , 4 1 , 5 1 , ……, 1 1 +n Since each term in the sequence is smaller than the term preceding it, the sequence is said to be a decreasing sequence. A sequence is decreasing if an + 1 < an for all n. 3. Find the 1st 5 terms of the sequence defined by an = n n 2 )1(− . Solution: Again by simple substitution, 1st term = a1 = 1 1 2 )1(− = - 2 1 2nd term = a2 = 2 2 2 )1(− = 4 1 3rd term = a3 = 3 3 2 )1(− = - 8 1 4th term = a4 = 4 4 2 )1(− = 16 1 7
8. 8. 5th term = a5 = 5 5 2 )1(− = - 32 1 The sequence defined by an = n n 2 )1(− can be written as - 2 1 , 4 1 , - 8 1 , 16 1 , - 32 1 ,…, n n 2 )1(− Notice that the presence of (-1) in the sequence has the effect of making successive terms alternately negative and positive. It is often useful to picture a sequence by sketching its graph w/ the n values as the x- coordinates and the an values as the y- coordinates. The corresponding graphs of Examples 1 - 3 are given below Example 1 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 n a n 8
9. 9. You can also find a specific term, given a rule for the sequence, as seen in the following example. 4. Find the 13th and 100th terms of the sequence whose general term is given by An = 2 )1( n n − Solution: For the 13th term, replace n with 13 and for the 100th term, replace n w/ 100: 13th term = a13 = 2 13 13 )1(− = 169 1− 100th term = a100 = 2 100 100 )1(− = 10000 1 Example 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0 1 2 3 4 5 n a n Example 3 -0.6 -0.4 -0.2 0 0.2 0.4 0 1 2 3 4 5 6 n a n 9
10. 10. Try this out A. Write the 1st 4 terms of the sequence whose nth term is given by the formula. 1. an = n + 1 2. an = 2 – 2n 3. an = n - 1 4. an = 2 n 5. an = 2n + 1 6. an = n² + 1 7. an = 3n - 1 8. an = 1+n n 9. an = 1 - 2n 10. an = n – n 1 11. an = 3n 12. an = (-1) n+1 n 13. an = n² - 1 14. an = (-1) n 2 n 15. an = n² - n 1 16. an = n n 12 − 17. an = 1 )1( 1 + − + n n 18. an = 1 )1( 2 1 + − + n n 19. an = 2 1 3 1 +       n 20. an = 3 1 n³ + 1 21. an = (1)n (n²+2n+1) B. Find the indicated term of the sequence whose nth term is given by the formula. 22. an = 3n + 4 a12 23. an= n(n -1) a11 24. an= (-1) n - 1 n² a15 25. an= ( 2 1 )n a8 26. an = 2n - 5 a10 27. an = 1+n n a12 28. an = (-1) n - 1 (n - 1) a25 29. an = ( 3 2 ) n a5 30. an = (n + 2)(n + 3) a17 10
11. 11. 31. an = (n + 4)(n + 1) a7 32. an = 2 12 )1( n n+ − a6 33. an = 4 )1( 2 + − n n a16 34. an = 2 3 n² - 2 a8 35. an = 3 1 n + n² a6 Lesson 3 Finding the nth Term of a Sequence In Lesson 2, some terms of a sequence were found after being given the general term. In this lesson, the reverse is done. That is, given some terms of the sequence, try to find the formula for the general term. Examples: 1. Find a formula for the nth term of the sequence 2, 8, 18, 32,… Solution: Solving a problem like this involve some guessing. Looking over the first 4 terms, see that each is twice a perfect square: 2 = 2(1) 8 = 2(4) 18 = 2(9) 32 = 2(16) By writing each sequence with an exponent of 2, the formula for the nth term becomes obvious: a1 = 2 = 2(1)² a2 = 8 = 2(2)² a3 = 18 = 2(3)² a4 = 32 = 2(4)² . . . an = 2(n)² = 2n² 11
12. 12. The general term of the sequence 2, 8, 18, 32,…. is an = 2n². 2. Find the general term for the sequence 2, 8 3 , 27 4 , 14 5 ,…. Solution: The first term can be written as 1 2 . The denominators are all perfect cubes while the numerators are all 1 more than the base of the cubes of the denominators: a1 = 2/1 = 3 1 11 + a2 = 3/8 = 3 2 12 + a3 = 4/27 = 2 3 13 + a4 = 5/64 = 3 4 14 + Observing this pattern, recognize the general term to be an = 3 1 n n + 3. Find the nth term of a sequence whose first several terms are given 2 1 , 4 3 , 6 5 , 8 7 , . . . Solution: Notice that the numerators of these fractions are the odd numbers and the denominators are the even numbers. Even numbers are in the form usually written in the form 2n, and odd numbers are written in the form 2n – 1 (an odd number differs form an even number by 1). So, a sequence that has these numbers for its first four terms is given by an = n n 2 12 − . 4. Find the nth term of a sequence whose first several terms are given -2, 4, -8, 16, -32,… Solution: 12
13. 13. These numbers are powers of 2 and they alternate in sign, so a sequence that agrees with these terms is given by an = (-1)n 2n . Note: Finding the nth term of a sequence from the 1st few terms is not always automatic. That is, it sometimes takes a while to recognize the pattern. Don’t be afraid to guess the formula for the general term. Many times an incorrect guess leads to the correct formula. Some pointers on how to find the general term of a sequence is given below. Pointers on How to Find the General Term of a Sequence 1. Study each term of the sequence as it compares to its term number. Then answer the following questions: a. Is it a multiple of the term number? b. Is it a multiple of the square or cube of the term number? If each term is a multiple of the term number, there will be a common number. 2. Examine the sequence. Does it increase or decrease? a. If it increases slowly, consider expressions that involve the term number plus or minus a constant like: n + 2 or n – 3. b. If it increases moderately, think about multiples of the term number plus or minus a constant like: 2n or 3n – 1. c. If the sequence increases very rapidly, try powers of the term number plus or minus a constant like: n2 or n2 + 1. 3. If the sequence consists of fractions, examine how the denominator and numerator change as separate sequences. For example: an = 2 1 n n + yields ,... 25 6 , 16 5 , 9 4 , 4 3 , 1 2 Also, though not all sequences can be defined by a formula, like for the sequence of prime numbers, be assured that the sequences discussed or given here are all obvious sequences that one can find a formula or rule for them. 13
14. 14. Try this out A. Write the formula for the nth term of the sequence: 1. The sequence of the natural numbers. 2. The sequence of the negative even integers. 3. The sequence of the odd natural numbers. 4. The sequence of the negative odd numbers. 5. The sequence of the multiples of 7. 6. The sequence of the positive even integers that are divisible by 4. 7. The sequence of the negative integers less than -5. 8. The sequence of the positive odd integers greater than 9. 9. The sequence for which a1 is 8 and each term is 6 more than the preceding term. 10. The sequence for which a1 is 3 and each term is 10 less than the preceding term. B. Determine the general term for each of the following sequences: 1. 2, 3, 4, 5,… 2. 3, 6, 9, 12,… 3. 4, 8, 12, 16, 20,… 4. 3, 4, 5, 6,… 5. 7, 10, 13, 16,… 6. 4, 9, 14, 19,… 7. 3, 12, 27, 48,… 8. 2, 16, 54, 128,… 9. 4, 8, 16, 32,… 10. 1, 8,27, 64,… 11. 1, 4, 9, 16,… Lesson 4 Recursively Defined Sequences Some sequences do not have simple defining formulas like those in Lesson 3. The nth term of a sequence may depend on some or all of the terms preceding it. A sequence defined in this way is called recursive. One has to be sure that the notations: an and an – 1 is understood first before going to the examples below. To illustrate: if the 24th term is an, then it can be written as a24. So that an – 1 means a24 – 1 = a23 or the 23rd term. 14 12. 3, 9, 27, 81,… 13. -2, 4, -8, 16,… 14. -3, 9, -27, 81,… 15. 32 1 , 16 1 , 8 1 , 4 1 ,… 16. 81 1 , 27 1 , 9 1 , 3 1 ,… 17. 25 4 , 16 3 , 9 2 , 4 1 ,… 18. 32 1 , 28 3 , 10 2 , 4 1 ,…
15. 15. Examples: 1. Find the first five terms of the sequence defined recursively by a1 = 1 and an = 3(an – 1 + 2). Solution: The defining formula asks that the preceding term, an – 1, be identified first before one can get the nth term, an. Thus, one can find the second term from the first term, the third term from the second term, the fourth term from the third term, and so on. Since the first term is given as a1 = 1, proceed as follows: a2 = 3(a1 + 2) = 3(1 + 2) = 9 a3 = 3(a2 + 2) = 3(9 + 2) = 33 a4 = 3(a3 + 2) = 3(33 + 2) = 105 a5 = 3(a4 + 2) = 3(105 + 2) = 321 Thus, the first five terms of this sequence are 1, 9, 33, 105, 321,… 2. Find the first 11 terms of the sequence defined recursively by F1 = 1, F2 = 1 and Fn = Fn – 1 + Fn – 2 Solution: To find Fn, the preceding terms, Fn – 1 and Fn – 2 have to be found first. Since the first two terms are given, then F3 = F2 + F1 = 1 + 1 = 2 F4 = F3 + F2 = 2 + 1 = 3 F5 = F4 + F3 = 3 + 2 = 5 By this time it should be clear as to what is happening here. Each term is simply the sum of the first two terms that precede it, so that its easy to write down as many terms as one pleases. Here are the first 11 terms: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,… The sequence in Example 2 is called the Fibonacci sequence, named after the 13th century Italian mathematician who used it to solve a problem about the breeding of rabbits. The sequence also occurs in a lot of other applications in nature. 15
16. 16. 8 5 3 2 1 1 The Fibonacci sequence In the branching of a tree To better understand the difference between recursively defined sequences and the general sequences encountered in the last three lessons, study the examples below. 3. Find the 120th term of the sequence defined by an = 2n + 3. Solution: Replacing n with 120, then 120th term = a120 = 2(120) + 3 = 243 The 120th term is 243. Notice that the 120th term from the example above is found without finding first the preceding term, that is, the 119th term! Whereas with the recursively defined sequence one has to find all the terms preceding the required term, 4. Find the 4th term of the recursively defined sequence an = 2 1−na where a1 = -3. Solution: First term = a1 = -3 Second term = a2 = 2 12−a = 2 1a = 2 3− Third term = a3 = 2 13−a = 2 2a = 2 2 3− = 4 3− Fourth term = a4 = 2 14−a = 2 3a = 2 4 3− = 8 3− The fourth term is - 8 3 . 16
17. 17. Try this out Find the first five terms of the given recursively defined sequences: 1. an = 2(an – 1 – 2) and a1 = 3 2. an = 2 1−na and a1 = -8 3. an = 2an – 1 + 1 and a1 = 1 4. an = 11 1 −+ na and a1 = 1 5. an = an – 1 + an – 2 and a1 = 1, a2 = 2 6. an = an – 1 + an – 2 + an – 3 and a1 = a2 = a3 = 1 Lesson 5 Arithmetic Sequences There are two main types of sequences. These are the arithmetic sequences and the geometric sequences. This lesson will show what arithmetic sequences are. The more detailed lesson on arithmetic sequences will be discussed in the next module. Look at the following sequences. 1. 4, 7, 10, 13,… 2. 33, 38, 43, 48,… 3. -2, -6, -10, -14,… 4. 100, 98, 96, 94,… 5. 2 1 , 1, 1 2 1 , 2, … Can you give the next two terms of the above sequences? How did you get the next terms? If you get 16 and 19 for a, then you are correct. Notice that a constant number , 3, is added to each term to get the next term. In b, 5 is added to the preceding term after the first, while in c, -4 is added to get the next term, in d, -2 is added to the preceding term and in e, ½ is added to get the next term. Notice that a constant or common number is added to the preceding term to get the next term in each of the sequences above. All these sequences are called 17
18. 18. arithmetic sequences. The constant number is called the common difference and is represented as d. To find the common difference, d, simply subtract the first term from the second term, a2 – a1, or the second term from the third term, a3 – a2, or the third term from the fourth term, a4 – a3; or in general, d = an – an – 1 Examples: 1. Determine if the sequence is arithmetic or not. If it is, find the common difference and the next three terms. Then graph. -11, -4, 3, 10,… Solution: To find out if the sequence is arithmetic, there must be a common difference between any two terms in the sequence. So that d = a2 – a1 = -4 – (-11) = 7 = a3 – a2 = 3 – (-4) = 7 = a4 – a3 = 10 – 3 = 7 The sequence is arithmetic and the common difference is 7. The next three terms are obtained by adding 7 to the preceding term, so that a5 = a4 + 7 = 10 + 7 = 17 a6 = a5 + 7 = 17 + 7 = 24 a7 = a6 + 7 = 24 + 7 = 31 Example 1 -20 -10 0 10 20 30 40 0 2 4 6 8 n an 18
19. 19. 2. Write the first five terms of the arithmetic sequence with first term 5 and common difference -2. Solution: The second term is found by adding -2 to the first term 5, getting 3. For the next term, add -2 to 3, and so on. The first five terms are 5, 3, 1, -1, -3. Remark: There is another way of finding the specified term of an arithmetic sequence but it will be discussed in the next module. The same thing is true for the general term of any arithmetic sequence. Try this out Determine whether the sequence is arithmetic or not. If it is, find the common difference and the next three terms. 1. 2, 5, 8, 11,… 2. 2, -4, 6, -8, 10,… 3. -6, -10, -14, -18,… 4. 40, 42, 44, 46,… 5. 1.2, 1.8, 2.4,… 6. 1, 5, 9, 13,… 7. ,... 5 1 , 4 1 , 3 1 , 2 1 8. ,...8,7,6,5 9. 98, 95, 92, 89,… 10. 1, 3 5 , 3 4 , 2,… Let’s Summarize 1. A sequence is a list of numbers in which order is important. a1, a2, a3, a4, …, an, … Each number in the list corresponds to each natural number. 2. A sequence may either be finite or infinite. A finite sequence has a specific number of terms. An infinite sequence has an endless number of terms. 19
20. 20. 3. To find the terms of a sequence given its rule, simply replace n with the number of the specific term needed to be found. 4. Most sequences have a general term or rule that describes all the terms in the sequence. There is no specific way of finding the general term of a given sequence. 5. A sequence is defined recursively when the nth term can be found only when the preceding term is found. 6. An arithmetic sequence is a sequence where each term is found by adding a constant number, called the common difference, to the preceding term. 7. The Fibonacci sequence is a special recursively defined sequence. What have you learned A. Write the first five terms of the sequence. 1. an = n n 7+ 3. an = 5n – 2 2. an = 3n - 1 B. Find the indicated term for the sequence. 4. an = -7n + 3; a8 6. an = 53 72 − + n n ; a14 5. an = (n + 2)(2n - 3); a5 C. Find the general term, an, for the given terms of the sequence. 7. 3, 7, 11, 15,… 9. ,... 16 1 , 8 1 , 4 1 , 2 1 8. 0, -4, -8, -12,… E. Find the first four terms of the sequence defined recursively. 10. a1= -1, an = 5an – 1 11. a1 = 6, an = 13 1 − − n an F. Find the common difference and the next three terms of the given arithmetic sequence. 12. 1, 10, 19, 28,… 14. 1, 3, 5, 7,… 13. 5.5, 7, 8.5, 10,… 15. 43, 39, 35, … 20
21. 21. 21
22. 22. Answer Key How much do you know 1 4, 5 8 , 4 7 ,2, 2 5 9. an = nn +2 1 2 1, 2, 4, 8, 16 10 -1, -3, -9, -27 3 1, 5, 9, 13, 17 11 5, 5, 6 5 , 2 5 4 -70 12 d = 6; 26, 32, 38 5 78 13 d = 1.5; 13, 14.5, 16 6 49/23 14 d = 1; 5, 6, 7 7 an = 4n 15 d = -3; 15, 12, 9 8 an = -10n Lesson 1 A. B. Because they can hold up traffic. Lesson 2 A. 1 2, 3, 4, 5 12 1, -2, 3, -4 2 0, -2, -4, -6 13 0, 3, 8, 15 3 0, 1, 2, 3 14 -2, 4, -8, 16 4 2, 4, 8, 16 15 0, 4 63 , 3 26 , 2 7 5 3, 5, 7, 9 16 0, 4 15 , 3 8 , 2 3 6 2, 5, 10, 17 17 5 1 , 4 1 , 3 1 , 2 1 −− 7 2, 5, 8, 11 18 17 1 , 10 1 , 5 1 , 2 1 −− 1 F 3 I 5 I 7 F 9 F 2 F 4 F 6 I 8 F 10 F 22
23. 23. 8 5 4 , 4 3 , 3 2 , 2 1 19 243 2 , 81 2 , 27 2 , 9 2 9 -1, -3, -5, -7 20 3 67 ,10, 3 11 , 3 4 10 0, 4 15 , 3 8 , 2 3 21 4, 9, 16, 25 11 3, 9, 27, 81 B. 22 40 29 243 32 23 110 30 380 24 225 31 88 25 256 1 32 36 1− 26 15 33 20 1 27 13 12 34 94 28 24 35 38 Lesson 3 A. B. 1 an = n + 1 10 an = n3 2 an = 3n 11 an = n2 3 an = 4n 12 an = 3n 4 an = n + 2 13 an = (-2)n 5 an = 3n + 4 14 an = (-3)n 6 an = 5n – 1 15 an = 1 2 1 +n 7 an = 3n2 16 an = n 3 1 8 an = 2n3 17 an = 2 )1( +n n 1 an = n 6 an = 4n 2 an = -2n 7 an = -(n + 5) 3 an = 2n – 1 8 an = 2n + 9 4 an = -2n + 1 9 an = 6n + 2 5 an = 7n 10 an = -10(n + 1)+ 3 23
24. 24. 9 an = 2n + 1 18 an = 13 +n n Lesson 4 1 3, 2, 0, -4, -12 4 1, 8 5 , 5 3 , 3 2 , 2 1 2 -8, -4, -2, -1, - 2 1 5 1, 2, 3, 5, 8 3 1, 3, 7, 15, 31 6 1, 1, 1, 3, 5 Lesson 5 1 Arithmetic d = 3 14, 17, 20 2 No 3 Arithmetic d = -4 -22, -26, -30 4 Arithmetic d = 2 48, 50, 52 5 Arithmetic d = 0.6 3, 3.6, 4.2 6 Arithmetic d = 4 17, 21, 25 7 No 8 No 9 Arithmetic d = 3 86, 83, 80 10 Arithmetic d = 1/3 3, 3 8 , 3 7 What have you learned 1 8, 5 12 , 4 11 , 3 10 , 2 9 9. an = n 2 1 2 1, 3, 9, 27, 81 10 -1, -5, -25, -125 3 3, 8, 13, 18, 23 11 6, 220 3 , 20 3 , 5 6 4 -53 12 d = 9; 37, 46, 55 5 49 13 d = 1.5; 11.5, 13, 14.5 6 35/37 14 d = 2; 9, 11, 13 7 an = 4n – 1 15 d = -4; 31, 27, 23 8 an = 4(1 – n) 24