Sequences, Series, and the Binomial Theorem Chapter 11 Sequences, Series, and the Binomial Theorem

§ 11.1 Sequences

Infinite and Finite Sequences An infinite sequence is a function whose domain is the set of natural numbers {1, 2, 3, 4,…}. 2, 4, 8, 16, 32,… terms An finite sequence is a function whose domain is the set of natural numbers {1, 2, 3, 4,…, n}, where n is some natural number. 2, 4, 6, 8, …, 98, 100 general term

Terms of a Sequences Example: Because a sequence is a function, it may be described as f(n) = an = 2n , where n is a natural number. 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.

Terms of a Sequences Example: 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

Terms of a Sequences Example: 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.

Applications Using Sequences Example: A mathematics lecture room has 20 rows with 8 seats in the first row, 12 seats in the second row, 16 seats in the third row and so on. Write an equation of a sequence whose term corresponds to the seats in each row. How many seats are there in the tenth row? Since each row has 4 more seats than the previous row, the general 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.

Arithmetic and Geometric Sequences § 11.2 Arithmetic and Geometric Sequences

Arithmetic Sequences In a sequence, when the difference of any two consecutive terms is a constant, the sequence is an arithmetic sequence. 5, 10, 15, 20, 25, … d = 5 The common difference is the constant, d.

Arithmetic Sequences Example: Write the first five terms of the arithmetic sequence whose first term is 3 and whose common difference is 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.

Arithmetic Sequences The general term, an, of an arithmetic sequence is given by an = a1 + (n – 1)d where a1 is the first term and d is the common difference. Example: 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

Arithmetic Sequences Example: Find the fifth term of an arithmetic sequence whose first three 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

Geometric Sequences A geometric sequence is a sequence in which each term (after the first) is obtained by multiplying the preceding term by a constant r. 5, 10, 20, 40, 80, … r = 2 The constant r is called the common ratio of the sequence.

Geometric Sequences Example: Write the first five terms of a geometric sequence whose first term is 2 and whose common ratio is 6. a1 = 2 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.

Geometric Sequences In 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

Geometric Sequences The general term, an, of a geometric sequence is given by an = a1rn – 1 where a1 is the first term and r is the common ratio. Example: Find the fifth term of the geometric sequence whose first term is 6 and whose common ratio is an = a1rn – 1

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, a9 = a1r9 – 1 = 3(– 4)8 = 196,608

§ 11.3 Series

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 a finite number of terms. Sequence Series 2, 4, 6, 8 2 + 4 + 6 + 8 = 20 5, 10, 20, 40 5 + 10 + 20 + 40 = 75

Infinite Series A series is an infinite series if it is the sum of all 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 used to mean “sum.”

Infinite Series The expression is read “the sum of 2n – 1 as n goes from 1 to 3.” This expression means the sum of the first three terms of the sequence whose general term is an = 2n – 1. Often the variable i is used instead of n. index of summation

Infinite Series Example: Evaluate

Partial Sums The sum of the first n terms of a sequence is a finite series known as a partial sum, Sn. S1 = a1 S2 = a1 + a2 S3 = a1 + a2 + a3 In general, Sn is the sum of the first n terms of a sequence.

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. an = a1 + (n – 1)d = 4 + (n – 1)6 = 4 + (n – 1)6 = 4 + 6n – 6 = 6n – 2

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 = 313n – 1 = 31 + (n – 1) = 3n

Partial Sums Example: Find the sum of the first four terms of the sequence whose general term is

Partial Sums of Arithmetic and Geometric Sequences § 11.4 Partial Sums of Arithmetic and Geometric Sequences

Partial Sums of Arithmetic Sequences The partial sum Sn of the first n terms of an arithmetic sequence is given by where a1 is the first term of the sequence and an is the nth term.

Partial Sums of Arithmetic Sequences Example: Find the sum of the first four terms of the arithmetic sequence 3, 9, 15, 21, 27, …

Partial Sums of Arithmetic Sequences Example: Find the sum of the first 25 even integers. Because 2, 4, 6, …, 50 is an arithmetic sequence, the formulas for Sn is used with n = 25, a1 = 2, and an = 50.

Partial Sums of Geometric Sequences The partial sum Sn of the first n terms of a geometric sequence is given by where a1 is the first term of the sequence, r is the common ratio, and r  1.

Partial Sums of Geometric Sequences Example: Find the sum of the first five terms of the geometric sequence 3, 12, 48, 192, 768, 3072, …

Partial Sums of Geometric Sequences Example: Chelsea made $20,000 during the first year she was self-employed. She made an additional 15% more than the previous 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 five years? 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.

Partial Sums of Geometric Sequences Example continued: b.) Chelsea made approximately $34,980.13 during her fifth year of self-employment, and a total of $134,853.33 during the first five years.

Infinite Geometric Sequences The sum S∞ of the terms of an infinite geometric sequence is given by where a1 is the first term of the sequence, r is the common ratio, and |r| < 1. If |r|  1, S does not exist.

Infinite Geometric Sequences Example: Find the sum of the terms of the geometric sequence

§ 11.5 The Binomial Theorem

Expanding Binomials Expanding a binomial such as (a + b)n means to write the factored form as a sum. (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 term 2 terms 3 terms 4 terms 5 terms 6 terms

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 The expansion of (a + b)n contains n + 1 terms. The first term is an and the last term is bn. The powers of a decrease by 1 for each term; the powers of b increase by 1 for each term. The sum of the exponents of a and b is n.

Pascal’s Triangle There are also patterns in the coefficients of the terms. When written in a triangular array, the coefficients are called Pascal’s triangle.

Pascal’s Triangle (a + b)0 (a + b)1 (a + b)2 (a + b)3 (a + b)4 1 1 1 2 1 1 3 3 1 4 6 4 1 1 5 10 10 5 1 n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 1 5 6 1 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.

Pascal’s Triangle Example: Expand (a + b)7. Use n = 7 row of Pascal’s triangle as the coefficients and 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

Factorials Factorial of n: n! If n is a natural number, then An alternative method for determining the coefficients 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.

Evaluating Factorials Example: Evaluate each expression.

Binomial Theorem Binomial Theorem If n is a positive integer, then It can be proved that the coefficients of terms in the expansion of (a + b)n can be expressed in terms of factorials. Following the earlier patterns and using the factorial expressions of the coefficients, we have the binomial theorem. Binomial Theorem If n is a positive integer, then

Binomial Theorem Example: Use the binomial theorem to expand (x + 3)4.

Binomial Theorem Example: Use the binomial theorem to expand (3a – 5b)6.

Binomial Expansion (r + 1)st Term in a Binomial Expansion The (r + 1)st term of the binomial expansion of (a + b)n is

Binomial Expansion Example: Find the ninth term in the expansion of (3x – 5y)10. n = 10, a = 3x, b = – 5y, r + 1 = 9, therefore r = 8