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Chapter 1

Chapter 1. Sequences and Series. 1.1 – Arithmetic sequences. Chapter 1: Sequences and Series. Create your own. Using the graph paper provided, create your own Fibonacci Spiral, like Vi Hart. Instructions (if you need them):

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Chapter 1

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  1. Chapter 1 Sequences and Series

  2. 1.1 – Arithmetic sequences Chapter 1: Sequences and Series

  3. Create your own • Using the graph paper provided, create your own Fibonacci Spiral, like Vi Hart. • Instructions (if you need them): • Begin in the right bottom third of the page, and draw a 1 cm x 1 cm square. • Below the first square, draw another 1 cm by 1 cm square. To the left of these squares, draw a 2 cm by 2 cm square. • Below these, draw a 3 cm by 3 cm square. • Carry on. • Use the diagonals between the corners to guide your curve.

  4. Notation • Say we wanted to represent the Fibonacci Sequence as follows: • 1 = t1 • 1 = t2 • 2 = t3 • Etc… • What would notation would we use for the 4th term? For the 7th? • What notation would we use for the nth term? • tn • How about the term before the nth term? • tn-1 • How about the term two before the nth term? • tn-2 • Can you make a general formula for tn in the Fibonacci Sequence? • tn = tn-1 + tn-2

  5. Some definitions A sequence is an ordered list of objects. It contains elements or terms that follow a pattern or rule to determine the next term in the in the sequence. The first term of the sequence is t1. The number of terms in the sequence is n. The general term of the sequence is tn. A finite sequence always has a finite number of terms. ex: 1, 2, 3, 4, 5 2, 4, 6, 8, …, 42 An infinite sequence never ends. ex: 2, 4, 6, 8, … An arithmetic sequence is a sequence in which there is a common difference between consecutive terms (i.e. the difference between terms is constant).

  6. challenge A grocer makes a display of cans in which the top row has one can and each subsequent row has two more cans than the row above it. If the display is made up of 100 cans, how many rows does it contain?

  7. Consider the sequence:10, 16, 22, 28, … • What’s the common difference? • What’s the next term? t1 = t2 = t3 = … tn= t1 t1 + d t1 + 2d t1 + (n-1)d

  8. Example A visual and performing arts group wants to hire a community events leader. The person will be paid $12 for the first hour of work, $19 for two hours of work, $26 for the three hours of work, and so on. a) Write the general term that you could use to determine the pay for any number of hours worked. b) What will the person get paid for 6h of work? • b) We are looking for the sixth term, t6. •  n = 6 •  t6 = 7(6) + 5 • t6 = 42 + 5 • t6 = 47 t1 = 12 t2 = 19 t3 = 26 • a) General term is always in the form: • tn = t1 + (n – 1)d • tn = 12 + (n – 1)7 • tn = 12 + 7n – 7 • tn = 7n + 5 What’s the common difference? 19 – 12 = 7 26 – 19 = 7 d = 7 They will get paid $47 for 6 hours of work. The general term is: tn= 7n +5

  9. On your calculator Let’s look at the same series: t1 = 12 t2 = 19 t3 = 26 tn= 7n + 5 The formula for the general term can be treated just like a function. So, we can graph the expression 7n+5 in our Y= on our graphing calculator, and use our graph to solve for n=6. Instructions: Y= 7 x + 5 GRAPH TRACE 6 You can see, to the right of your screen, it says y = 47, which was our answer. Now, if you wanted to find any term, you could just hit TRACE and the NUMBER, and it will give you your answer in the “y=“

  10. example The musk-ox of northern Canada are hoofed mammals that survived the Pleistocene Era, which ended 10 000 years ago. In 1955, the Banks Island musk-ox population was approximately 9250 animals. Suppose that in subsequent years, the growth of the musk-ox population followed an arithmetic sequence, in which the number of musk-ox increased by approximately 1650 each year. How many years would it take for the musk-ox population to reach 100 000? • tn = t1 + (n – 1)d • 100 000 = 9250 + (n – 1)1650 • 100 000 = 9250 + 1650n – 1650 • 100 000 – 9250 + 1650 = 1650n • 92 400 = 1650n • 56 = n What variable will 100 000 represent? What are we looking for? 9250, 10 900, 12 550, …, 100 000 t1 = 9250 d = 1650 tn = 100 000 n = ? It would take 56 years for the musk-ox population to reach 100 000.

  11. Try it! t3 = 16 t8 = 6 t9 = 4 a) d = 4 – 6 = -2 t1 = 16 + 2 + 2 t1 = 20 c) We’re told that the 8th row has six boxes, so we can see that’s third from the top. So, there must be 10 rows. Algebraically: 2 = —2n + 22  —20 = —2n  n = 10 • b) tn = t1 + (n – 1)(d) • tn= 20 + (n – 1)(-2) • tn = 20 – 2n + 2 • tn = —2n + 22

  12. Pg. 16–20 #4, 5, 8, 12, 14, 17, 20, 21, 23, 24 Independent practice

  13. 1.2 – Arithmetic Series Chapter 1: Sequences and Series

  14. Handout The handout follows the activity outlined on page 22 of your textbook. Answer all the questions to your fullest ability, as this is a summative assessment.

  15. Arithmetic series An arithmetic series is the sum of terms that form an arithmetic sequence. For the arithmetic sequence 2, 4, 6, 8, the arithmetic series is represented by 2 + 4 + 6 + 8. Recall: t1 = t2 = t3 = … tn = t1 t1 + d t1 + 2d t1 + (n-1)d We can express an arithmetic series as follows: Sn = t1 + (t1 + d) + (t1 + 2d) + … + (t1 + (n – 3)d) + (t1 + (n – 2)d) + (t1 + (n-1)d) How can we use Gauss’ method to create a formula for the sum of an arithmetic series?

  16. formula We can express an arithmetic series as follows: Sn = t1 + (t1 + d) + (t1 + 2d) + … + (t1 + (n – 3)d) + (t1 + (n – 2)d) + (t1 + (n-1)d) Sn = t1 + (t1 + d) + … + (t1 + (n – 2)d) + (t1 + (n-1)d) Sn = [t1+ (n-1)d] + [t1+ (n – 2)d] + … + (t1 + d) + t1 • 2Sn = [ 2t1 + (n – 1)d] + [2t1 + (n – 1)d] + … + [2t1 + (n – 1)d] + [2t1 + (n – 1)d] • 2Sn = n[2t1 + (n – 1)d] •  Sn = (n/2)[2t1 + (n – 1)d]

  17. Formula continued Alternatively, we can substitute tn into the formula to find: So, the two formulas for the sum of arithmetic series are:

  18. example The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Determine the first six terms of the series and the sum to six terms. Given: S2= 13 S4 = 46 So, the first six terms are 4, 9, 14, 19, 24, 29. and S6 = (6/2)(4+29) = 3(33) = 99 Solve the system of equations: Substitute d = 5 into one of the equations: 1 2 1 – 2

  19. Try it! The sum of the first two terms of an arithmetic series is 19 and the sum of the first four terms is 50. What are the first six terms of the series and the sum to 20 terms? Given: S2 = 19 S4 = 50 System of Equations: The first six terms are 8, 11, 14, 17, 20, 23. S20= (20/2)(2(8) + (20-1)3) = 10(73) = 730

  20. Pg. 27 – 31, # 1, 6, 9, 12, 14, 16-20, 23 Independent practice

  21. 1.3 – Geometric sequences Chapter 1: Sequences and Series

  22. Geometric sequences A geometric sequences is a sequence in which the ratio of consecutive terms is constant. The Fibonacci Sequence is an example of a geometric sequence. The common ratio of a geometric sequence is the ratio of successive terms in that sequence. Do the activity described on page 33. Answer questions # 1 – 4, and create a table like that seen in question #2.

  23. Table from coin toss activity

  24. Formula We can always find r, the common ratio, by dividing one term by the one preceding it, or by using this formula: The general term for a geometric sequence where n is a positive integer is: tn = t1rn-1 Terms of a geometric sequence: t1 = t1 t2 = t1r t3 = t1r2 t4 = t1r3 … tn = t1rn-1

  25. example Sometimes you use a photocopier to create enlargements or reductions. Suppose the actual length of a photograph is 25 cm and the smallest size that copier can make is 67% of the original. What is the shortest possible length of the photograph after 5 reductions? Express your answer to the nearest tenth of a centimetre. • tn = t1rn-1 • t6 = (25)(0.67)5 • t6= 3.375… t1 = r = n = 25 cm 0.67 6 After five reductions, the shortest possible length of the photograph is approximately 3.4 cm.

  26. EXAMPLE In a geometric sequence, the third term is 54 and the sixth term is -1458. Determine the values of t1 and r, and list the first three terms of the sequence. t2 = 54/r = 54/-3 = -18 t1 = -18/-3 = 6 t1 = 6 Method 1: Method 2: t3 = 54 t6 = -1458 • tn = t1rn-1 • 54 = t1r2 • t1 = 54/r2 • Substitute: •  -1458 = t1r5 •  r5 = -1458/(54/r2) • r5 = -1458(r2/54) • r5/r2 = -27 • r3 = -27 • r = -3 • From there it’s the same! t3 = 54 t4 = 54r t5 = 54r2 t6 = 54r3 = -1458 The first three terms are 6, -18, 54. t6 = 54r3 = -1458 r3 = -1458/54 = -27 r = -3

  27. example The modern piano has 88 keys. The frequency of the notes ranges from A0, the lowest note, at 27.5 Hz, to C8, the highest note on the piano, at 4186.009 Hz. The frequencies of these notes approximate a geometric sequence as you move up the keyboard. Determine the common ratio of the geometric sequence produced from the lowest key, A0, to the fourth key, C1, at 32.7 Hz. Use the lowest and highest frequencies to verify the common ratio found in part a). • t1 = 27.5 Hz • n = 4 • tn = 32.7 Hz • tn = t1rn-1 • 32.7 = (27.5)r3 • r3 = 1.189… • r = 1.0594… • The common ratio is approximately 1.06. • b) t88 = t1r88-1 • 4186.009 = (27.5)r87 • r87 = 152.218… • r = 1.0594… • It’s the same! Do you know how to find the 87th root of a number on your calculator?

  28. Pg. 39–45, # 1, 5, 9, 10, 13, 15, 16, 19, 21, 23, 24, 27 Independent practice

  29. 1.4 – Geometric series Chapter 1: Sequences and Series

  30. Geometric series Considering what you know about arithmetic series, do you think a geometric series might be? A geometric series is the terms of a geometric sequence expressed as a sum. The formula for the sum of a geometric series is:

  31. example Determine the sum of the first 10 terms of each geometric series. 4 + 12 + 36 + … t1 = 5, r = 1/2 a) In the series, t1 = 4, r = 3, and n = 10. b) t1 = 5, r = ½, n = 10

  32. example Determine the sum of each geometric series. 1/27 + 1/9 + 1/3 + … + 729 tn = t1rn-1 729 = (1/27)(3)n-1 (27)(729)=3n-1 (33)(36) = 3n-1 39 = 3n-1 9 = n – 1 n = 10 There are ten terms in this series. a) The sum of the series is 29524/27.

  33. Alternate Formula method Determine the sum of each geometric series. 1/27 + 1/9 + 1/3 + … + 729 We can make an alternate formula that can make these kind of problems one step! The general formula can also be written like this. Using the general formula for a geometric sequence, we can multiply each side by r to find that t1rn can be equal to rtn Substitute: Our series:

  34. Revised formula

  35. example The Western Scrabble Network is an organization whose goal is to promote the game of Scrabble. It offers Internet tournaments throughout the year that WSN members participate in. The format of these tournaments is such that the losers of each round are eliminated from the next round. The winners continue to play until a final match determines the champion. If there are 256 entries in an Internet Scrabble tournament, what is the total number of matches that will be played in the tournament? The players match up and play against each other, so for the first round there are 256/2 = 128 matches = t1. Each round, half the players are eliminated, so r = ½. At the end of the tournament there will be only one final match, so tn = 1. There will be 255 matches played at the tournament.

  36. Pg. 53–57, # 4, 6, 8, 10, 11, 13, 15, 16-19, 22 Independent practice

  37. 1.5 – infinite geometric series Chapter 1: Sequences and Series

  38. Zeno’s paradox

  39. Pull out a sheet of paper Draw a line dividing your page in two halves (fold it if necessary). Shade in one half. Divide the non-shaded half into two, and shade one of those halves. Divide the new non-shaded half into two, and shade one of those halves. Continue on at least six more times. What’s the fraction of each part? So, the first five terms are: 1/2, 1/4, 1/8, 1/16, 1/32 … 1/32 1/8 1/16 What will the next two terms be? What’s the general term? 1/2 1/4 tn = (1/2)(1/2)n-1 = (1/2)n

  40. Graph it • Enter (1/2)x into your Y= on your calculator. • What does the graph look like? • What happens to the y-value as the x-value gets larger? tn = (1/2)n What will our the sum of this geometric series look like? • Enter 1 – (1/2)x into your Y= • What does the graph look like? Look at the table of values. • What happens as the x-value gets larger? What’s it getting closer to?

  41. Vi Hart on Infinite Series More?

  42. Convergent series Consider the series 4 + 2 + 1 + ½ + … A series is a convergent series when it has an infinite number of terms, in which the sequence of partial sums approaches a fixed value. We can tell this series is convergent by looking at its graph and table of values, and seeing that it is approaching the value of 8. Graph it! Put —8[(1/2)x – 1] into your Y= and check the graph.

  43. Divergent series Consider the series 4 + 8 + 16 + 32 + … A series is a divergent series when it has an infinite number of terms, in which the sequence of partial sums does not approach a fixed value. We can see from looking at this graph and the table of values that the y-value is not approaching any number—it’s just getting bigger. Graph it! Put 4(2x – 1) into your Y= and check the graph.

  44. formula We can re-arrange for the formula for a finite geometric series to find: Why might this be helpful when r < 1? So, if r < 1, then as n gets very large, what will happen to rn? rn gets closer to zero as n gets larger for r < 1. So, for an infinite series, where n is getting larger and larger, approaching infinity, what can we say about (1 – rn)? What can we say about Sn? For an infinite series: where -1 < r < 1

  45. example • Assume that each shaded square represents ¼ of the area of the larger square bordering two of its adjacent sides and that the shading continues indefinitely in the indicated manner. • Write the series of terms that would represent this situation. • How much of the total area of the largest square is shaded? What portion of the square does each shaded region represent?  1/4 + 1/16 + 1/64 + … 1/64 1/16 b) The total area shaded is 1/3 of the largest square. 1/4

  46. P. 63-65 #1, 5, 6, 8, 10, 12, 14, 16, 18, 19, 20, 21, 22 Independent practice

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