Sequences, Series, and the Binomial Theorem

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Sequences, Series, and the Binomial Theorem

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Sequences, Series, and the Binomial Theorem

  1. 1. Infinite and Finite SequencesAn infinite sequence is a function whose domainis the set of natural numbers {1, 2, 3, 4,…}. 2, 4, 8, 16, 32,… termsAn finite sequence is a function whose domainis the set of natural numbers {1, 2, 3, 4,…, n},where n is some natural number. 2, 4, 6, 8, …, 98, 100 general term
  2. 2. Terms of a SequencesBecause a sequence is a function, it may bedescribed as f(n) = an = 2n , where n is a naturalnumber. Example: Write the first three terms of the sequence whose general term is given by an = n3 + 1. a1 = 13 + 1 = 1 + 1 = 2 Replace n with 1. a2 = 23 + 1 = 8 + 1 = 9 Replace n with 2. a3 = 33 + 1 = 27 + 1 = 28 Replace n with 3. The first three terms are 2, 9, 28, and 65.
  3. 3. Terms of a SequencesExample:If the general term of a sequence is given by an = 4n2+ 5, find a.) a10, b.) the two-hundredth term of the sequence. a.) a10 = 4(10)2 + 5 Replace n with 10. = 4(100) + 5 = 405 b.) a200 = 4(200)2 + 5 Replace n with 200. = 4(40000) + 5 = 160005
  4. 4. Terms of a SequencesExample: Find a general term an of the sequence whose first five terms are a.) 5, 10, 15, 20, 25; b.) – 1, 2, 7, 14, 23. a.) These numbers are the product of 5 and the first five natural numbers, so a general term might be an = 5n. b.) These numbers are 2 less than the squares of the first five natural numbers, so a general term might be an = n2 – 2.
  5. 5. Applications Using SequencesExample:A mathematics lecture room has 20 rows with 8 seats inthe first row, 12 seats in the second row, 16 seats in thethird row and so on. Write an equation of a sequencewhose term corresponds to the seats in each row. Howmany seats are there in the tenth row?Since each row has 4 more seats than the previous row, thegeneral term would be an = 8 + 4(n – 1) where n is 1, 2, 3, …,20.To find the number of seats in the tenth row,evaluate a10 = 8 + 4(10 – 1) = 8 + 4(9) = 8 + 36 = 44.
  6. 6. Arithmetic SequencesIn a sequence, when the difference of any twoconsecutive terms is a constant, the sequence is anarithmetic sequence. 5, 10, 15, 20, 25, … d=5 The common difference is the constant, d.
  7. 7. Arithmetic SequencesExample:Write the first five terms of the arithmetic sequencewhose first term is 3 and whose common differenceis 4. a1 = 3 a2 = 3 + 4 = 7 a3 = 7 + 4 = 11 a4 = 11 + 4 = 15 a5 = 15 + 4 = 19 The first five terms are 3, 7, 11, 15, and 19.
  8. 8. Arithmetic SequencesThe general term, an, of an arithmetic sequence is givenby Example: a = a + (n – 1)d n 1where a1 is the first term and d is the commondifference. Consider the arithmetic sequence whose first term is 3 and common difference is 4. Write an expression for the general term an. an = a1 + (n – 1)d = 3 + (n – 1)4 = 3 + 4n – 4 = 4n – 1
  9. 9. Arithmetic SequencesExample:Find the fifth term of an arithmetic sequence whose firstthree terms are 6, 11, 16. The fifth term of the sequence is a5 = a1 + (5 – 1)d = a1 + 4d. a1 is the first term of the sequence, so a1 = 6. d is the common difference of terms, so d = a2 – a1 = 11 – 6 = 5. Thus, a5 = a1 + 4d = 6 + 4(5) = 6 + 20 = 26
  10. 10. Geometric SequencesA geometric sequence is a sequence in which eachterm (after the first) is obtained by multiplying thepreceding term by a constant r. 5, 10, 20, 40, 80, … r=2 The constant r is called the common ratio of the sequence.
  11. 11. Geometric SequencesExample: Write the first five terms of a geometric sequence whose first term is 2 and whose common 2 a = ratio is 6. 1 a2 = 2(6) = 12 a3 = 12(6) = 72 a4 = 72(6) = 432 a5 = 432(6) = 2592 The first five terms are 2, 12, 72, 432, and 2592.
  12. 12. Geometric SequencesIn the geometric sequence whose first five terms are 2, 12, 72,432, and 2592, notice the general pattern of the terms. a1 = 2 a2 = 2(6) = 12 or a2 = a1(r) a3 = 12(6) = 72 or a3 = a2(r) = (a1 ∙ r) ∙ r = a1r2 a4 = 72(6) = 432 or a4 = a3(r) = (a1 ∙ r2) ∙ r = a1r3 a5 = 432(6) = 2592 or a5 = a4(r) = (a1 ∙ r3) ∙ r = a1r4 (subscript – 1) is the power
  13. 13. Geometric SequencesThe general term, an, of a geometric sequence is given by an = a1 rn – 1 where a1 is the first term and r is the common ratio. Example: Find the fifth term of the geometric sequence 1 whose first term is 6 and whose common ratio is . n 1 2 1 an = a1rn – 1 6 2 5 1 4 1 1 1 6 3 a5 6 6 6 2 2 16 16 8
  14. 14. Geometric Sequences Example: Find the ninth term of the geometric sequence whose first three terms are 3, –12, 48. Since r is the common ratio of the terms, a2 12r 4. a1 3 a9 = a1r9 – 1 = 3(– 4)8 = 196,608
  15. 15. Geometric means are the terms between any twononconsecutive terms of a geometric sequence.
  16. 16. Example 4: Finding Geometric MeansFind the geometric mean of and . Use the formula.
  17. 17. Check It Out! Example 4Find the geometric mean of 16 and 25. Use the formula.
  18. 18. A Harmonic Progression is asequence of quantities whose reciprocals form an arithmetic progression.
  19. 19. * The series formed by thereciprocals of the terms of a geometric series are also geometric series.
  20. 20. * There is no general method of finding the sum of a harmonic progression.
  21. 21. ExampleThe Sequence“s1 , s2 , … , sn”is a Harmonic Progression if“1/s1 , 1/s2 , … , 1/sn”forms an Arithmetic Progression.
  22. 22. Method ForRe-checking a Harmonic Progression
  23. 23. A Harmonic Progression is a set of values that, once reciprocated, resultsto an Arithmetic Progression. To check ,the reciprocated values must possess arational common difference. Once this has been identified, we may say that the sequence is a Harmonic Progression.
  24. 24. Harmonic Means are theterms found in betweentwo terms of a harmonic progression.
  25. 25. Determine which of thefollowing are Harmonic Progressions.
  26. 26. 1) 1 ,1/2 , 1/3 , 1/4 , ...
  27. 27. Step 1: Reciprocate all the given terms.* The reciprocals are: 1 , 2 , 3 , 4 , …Step 2: Identify whether the reciprocatedsequence is an Arithmetic Progression by checkingif a common difference exists in the terms.
  28. 28. Answer: It is a Harmonic Progression.
  29. 29. 2) 1 , 1/4 , 1/5 , 1/7 , ...
  30. 30. Step 1: Reciprocate all the given terms.* The reciprocals are: 1 , 4 , 5 , 7 , …Step 2: Identify whether the reciprocatedsequence is an Arithmetic Progression by checking if a common difference exists inthe terms.
  31. 31. Answer: It is NOT aHarmonic Progression.
  32. 32. Determine the next three terms of each of the following Harmonic Progressions.
  33. 33. 1) 24 , 12 , 8 , 6 , …
  34. 34. Solution: 24 , 12 , 8 , 6 , … = 1/24 , 1/12 , 1/8 , 1/6* To find the common difference: 1/12 – 1/24 = 2/24 – 1/24 = 1/24
  35. 35. You can subtract the second term to the first term, the third to thesecond term, the forth tothe third term, and so on and so forth.
  36. 36. To get the next threeterms: 5th Term = 1/6 + 1/24 = 4/24 + 1/24 = 5/24 * Reciprocate = 24/5
  37. 37. 6th Term = 5/24 + 1/24 = 6/24 = 1/4 * Reciprocate =4
  38. 38. 7th Term = 1/4 + 1/24 = 6/24 + 1/24 = 7/24 * Reciprocate = 24/7
  39. 39. Find the Harmonic Mean between the following terms.
  40. 40. 1) 12 and 8
  41. 41. Step 1: Reciprocate all thegiven terms.* The reciprocals are: 1/12and 1/8Step 2: Arrange the giventerms as follows:
  42. 42. 1/12 Harmonic Mean 1/81’st term 2’nd term 3’rd term
  43. 43. *For this problem, we will use the formula: tn = t1 + (n – 1)d
  44. 44. We may now substitutethe values in the problemto the formula to find the common difference (d) and the Harmonic Mean as follows:
  45. 45. t3 = t1 + (3 - 1)d 1/8 = 1/12 + 2d 1/8 – 1/12 = 2d(3 – 2) / 24 = 2d (3 – 2) = 48d 1 = 48d d = 1/48
  46. 46. *After getting theCommon Difference, add it to the first term to get the Harmonic Mean between the two terms.
  47. 47. t2 = t1 + d= 1/12 + 1/48= (4 + 1) / 48 = 5/48*Reciprocate = 48/5
  48. 48. Insert three Harmonic Means between the following terms:
  49. 49. 1) 36 and 36/5
  50. 50. Step 1: Reciprocate all thegiven terms.* The reciprocals are: 1/36and 5/36Step 2: Arrange the giventerms as follows:
  51. 51. 1’st term Harmonic Means2’nd , 3’rd , and 4’th term 5/36 5’th term
  52. 52. *For this problem, we will use the formula: tn = t1 + (n – 1)d
  53. 53. We may now substitutethe values in the problemto the formula to find the common difference (d)and the Harmonic Means as follows:
  54. 54. t5 = t1 + (5 - 1)d5/36 = 1/36 + 4d5/36 – 1/36 = 4d(5 - 1) / 36 = 4d (5 - 1) = 144d 4 = 144d d = 4/144 = 1/36
  55. 55. *After getting theCommon Difference, add it to the first term, thenadd it to the second term, and then add it to the third term to get theHarmonic Means between the two terms.
  56. 56. t2 = t1 + d= 1/36 + 1/36 = 2/36 = 1/18*Reciprocate = 18
  57. 57. t3 = t2 + d= 2/36 + 1/36 = 3/36 = 1/12*Reciprocate = 12
  58. 58. t4 = t3 + d= 3/36 + 1/36 = 4/36 = 1/9*Reciprocate =9
  59. 59. Therefore, the threemeans between 36 and 36/5 are 18, 12, and 9.
  60. 60. Activity
  61. 61. Determine if the followingare harmonicprogressions or not:1) 1/12 , 1/24 , 1/362) 2 , 5 , 7 , 83)1/5 , 1/10 , 1/15
  62. 62. Find the next three termsin the following harmonicprogressions:1) 1/2 , 1/5 , 1/8 , 1/11 , …2) 19 , 17 , 15 , 13, …3) 12 , 6 , 4 , 3 , …
  63. 63. Find the harmonic meanbetween:1) 1/2 and 1/52) 1 and 1/9
  64. 64. Insert three harmonicmeans between:1) 1/2 and 1/82) 1 and 1/10
  65. 65. Finite Series A sum of the terms of a sequence is called a series.A series is a finite series if it is the sum of afinite number of terms. Sequence Series 2, 4, 6, 8 2 + 4 + 6 + 8 = 20 5, 10, 20, 40 5 + 10 + 20 + 40 = 75
  66. 66. Infinite SeriesA series is an infinite series if it is the sum ofall the terms of the sequence. Sequence Series 2, 4, 6, 8, … 2+4+6+8+… 5, 10, 20, 40, … 5 + 10 + 20 + 40 + …When the general term of a sequence is known,summation notation is used for denoting a series.The Greek uppercase letter sigma, Σ, is usedto mean “sum.”
  67. 67. Infinite Series 3The expression (2n 1) is read “the sum of 2n – n 11 as n goes from 1 to 3.” This expression meansthe sum of the first three terms of the sequencewhose general term is an = 2n – 1.Often the variable i is used instead of n. 3 (2i 1) (2 1 1) (2 2 1) (2 3 1) i 1 index of 1 3 5summation 9
  68. 68. Infinite SeriesExample: 5Evaluate (n2 2). i 0 5 (n 2 2) 02 2 12 2 22 2 32 2 i 0 2 2 4 2 5 2 2 3 6 11 18 27 67
  69. 69. Partial SumsThe sum of the first n terms of a sequence isa finite series known as a partial sum, Sn. S1 = a1 S2 = a1 + a2 S3 = a1 + a2 + a3In general, Sn is the sum of the first n terms of asequence. n Sn an i 1
  70. 70. Partial Sums Example: Write the series using summation notation. 4 + 10 + 16 + 22 + 28 Since the difference of each term and the preceding term is 6, this is an arithmetic sequence with a1 = 4 and d = 6. a + (n – 1)d = 4 + (n – 1)6 a = n 1 = 4 + (n – 1)6 = 4 + 6n – 6 = 6n – 2 54 + 10 + 16 + 22 + 28 6i 2 i 1
  71. 71. Partial Sums Example: Write the series using summation notation. 3 + 9 + 27 + 81 + 243 Since each term is the product of the preceding term and 3, this is a geometric sequence with a1 = 3 and r = 3. an = a1rn – 1 = 3(3)n – 1 = 3 13 n – 1 = 31 + (n – 1) = 3n 5 i3 + 9 + 27 + 81 + 243 3 i 1
  72. 72. Example:Find the sum of the first four terms of the sequence 5 nwhose general term is n a . 3n 45 i 5 1 5 2 5 3 5 4 S4 i 1 3i 31 3 2 3 3 3 4 4 3 2 1 3 6 9 12 48 18 8 3 36 36 36 36 77 5 2 36 36
  73. 73. Partial Sums of Arithmetic SequencesThe partial sum Sn of the first n terms of an arithmetic sequenceis given by n Sn (a1 an) 2where a1 is the first term of the sequence and an is the nth term.
  74. 74. Example:Find the sum of the first four terms of thearithmetic sequence 3, 9, 15, 21, 27, … n Sn (a1 an) 2 4 S4 (3 21) 2 2(24) 48
  75. 75. Example:Find the sum of the first 25 even integers.Because 2, 4, 6, …, 50 is an arithmetic sequence, theformulas for Sn is used with n = 25, a1 = 2, and an = 50. n Sn (a1 an) 2 25 S25 (2 50) 2 25 (52) 2 650
  76. 76. The partial sum Sn of the first n terms of ageometric sequence is given by a1(1 r n) Sn 1 rwhere a1 is the first term of the sequence, r isthe common ratio, and r 1.
  77. 77. Example:Find the sum of the first five terms of thegeometric sequence 3, 12, 48, 192, 768, 3072, … a1(1 r n) Sn 1 r 12 3(1 45) r 4 S5 3 1 4 3(1 1024) 3 3( 1023) 3 1023
  78. 78. Example:Chelsea made P20,000 during the first year she was self-employed. She made an additional 15% more than theprevious year in each subsequent year.a.) How much did she make during her fifth year of business?b.) What were her total earnings during the fiveyears? Chelsea’s earnings are modeled by a geometric sequence where n = 5, a1 = 20,000, and r = 1.15 a.) an = a1rn – 1 a5 = 20,000(1.15)4 34,980.13 Continued.
  79. 79. Example continued: b.) S a1(1 r n) n 1 r 20,000(1 1.155) S5 1 1.15 20,000(1 2.0114) 0.15 20,000( 1.0114) 134,853.33 0.15Chelsea made approximately P34,980.13 during herfifth year of self-employment, and a total ofP134,853.33 during the first five years.
  80. 80. Infinite Geometric SequencesThe sum S∞ of the terms of an infinite geometricsequence is given by a1 S 1 r where a1 is the first term of the sequence, r is the common ratio, and |r| < 1. If |r| 1, S does not exist.
  81. 81. Example: Find the sum of the terms of the geometric sequence 4, 8 , 16 , 32 , 3 9 27 a1 S 1 r 8 2r 4 4 3 3 2 1 3 4 12 1 3
  82. 82. Expanding BinomialsExpanding a binomial such as (a + b)n meansto write the factored form as a sum.(a + b)0 = 1 1 term(a + b)1 = a + b 2 terms(a + b)2 = a2 + 2ab + b2 3 terms(a + b)3 = a3 + 3a2b + 3ab2 + b3 4 terms(a + b)4 = a4 + 4a3b + 6a2b2 + 4a1b3 + b4 5 terms(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 6 terms
  83. 83. Expanding Binomials(a + b)0 = 1(a + b)1 = a + b(a + b)2 = a2 + 2ab + b2(a + b)3 = a3 + 3a2b + 3ab2 + b3(a + b)4 = a4 + 4a3b + 6a2b2 + 4a1b3 + b4(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 1. The expansion of (a + b)n contains n + 1 terms. 2. The first term is an and the last term is bn. 3. The powers of a decrease by 1 for each term; the powers of b increase by 1 for each term. 4. The sum of the exponents of a and b is n.
  84. 84. Pascal’s TriangleThere are also patterns in the coefficients of the terms.When written in a triangular array, the coefficients arecalled Pascal’s triangle.
  85. 85. Pascal’s Triangle(a + b)0 1 n=0(a + b)1 1 1 n=1(a + b)2 1 2 1 n=2(a + b)3 1 3 3 1 n=3(a + b)4 1 4 6 4 1 n=4(a + b)5 1 1 5 10 10 5 1 n=5 1 6 6 15 20 15 6 1 Add the consecutive numbers in the row for n = 5 and write each sum “between and below” the pair.
  86. 86. Pascal’s Triangle Example: Expand (a + b)7.Use n = 7 row of Pascal’s triangle as the coefficientsand the noted patterns. 1 6 15 20 15 6 1 n=6 1 7 21 35 35 21 7 1 n=7 (a + b)7 = 1a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 + 1b7
  87. 87. FactorialsAn alternative method for determining thecoefficients of (a + b)n is based on using factorials. The factorial of n, written n! (read “n factorial”), is the product of the first n consecutive natural numbers. Factorial of n: n! If n is a natural number, then n! = n(n – 1)(n – 2)(n – 3) . . . ∙ 3 ∙ 2 ∙ 1. The factorial of 0, written 0!, is defined to be 1.
  88. 88. Evaluating FactorialsExample:Evaluate each expression. a.) 6! 8! b.) 3! a.) 6! 6 5 4 3 2 1 720 8! 8 7 6 5 4 3 2 1 b.) 6720 3! 3 21
  89. 89. Binomial TheoremIt can be proved that the coefficients of terms in the expansion of (a+ b)n can be expressed in terms of factorials. Following the earlierpatterns and using the factorial expressions of the coefficients, wehave the binomial theorem. Binomial Theorem If n is a positive integer, then n n 1 1 n(n 1) n 2 2 (a b) n an a b a b 1! 2! n(n 1)(n 2) n 3 3 a b  bn 3!
  90. 90. Binomial TheoremExample:Use the binomial theorem to expand (x + 3)4. 4 4 4 3 4 3 2 2 4 3 2 3 4 (x 3) x x3 x3 x3 3 1! 2! 3! x4 4 3x3 6 9x2 4 27x 81 x4 12x3 54x2 108x 81
  91. 91. Binomial TheoremExample:Use the binomial theorem to expand (3a – 5b)6. 6 6 6 5 6 5 4 2 (3a 5b) (3a) (3a) ( 5b) (3a) ( 5b) 1! 2! 6 5 4 6 5 4 3 (3a)3( 5b)3 (3a) 2( 5b) 4 3! 4! 6 5 4 3 2 (3a)( 5b)5 ( 5b)6 5! 729a 6 7290a5b 30,375a 4b 2 67,500a 3b3 84,375a 2b 4 56, 250ab5 15,625b6
  92. 92. Binomial Expansion(r + 1)st Term in a Binomial ExpansionThe (r + 1)st term of the binomial expansion of n!(a + b)n is a n rb r. r !(n r)!
  93. 93. Binomial ExpansionExample:Find the ninth term in the expansion of (3x – 5y)10. n = 10, a = 3x, b = – 5y, r + 1 = 9, therefore r = 8 n! 10! a n rbr (3x)10 8( 5 y)8r !(n r)! 8!(10 8)! 45 9x 2 390,625 y 8 158, 203,125x 2 y 8

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