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Chapter 6 Sequences And Series Look at these number sequences carefully can you guess the next 2 numbers? What about guess the rule? +10 30 40 50 60 70 80 --------------------------------------------------------------------------------------------------------------------- 17 20 23 26 29 32 +3 --------------------------------------------------------------------------------------------------------------------- 48 41 34 27 20 13 -7 Can you work out the missing numbers in each of these sequences? 50 75 100 125 150 175 +25 --------------------------------------------------------------------------------------------------------------------- 30 50 70 90 110 130 +20 --------------------------------------------------------------------------------------------------------------------- 196 191 186 181 176 171 -5 --------------------------------------------------------------------------------------------------------------------- 306 296 286 276 266 256 -10 Now try these sequences – think carefully and guess the last number! 1 2 4 7 11 +1, +2, +3 … 16 --------------------------------------------------------------------------------------------------------------------- 3 6 12 24 48 96 double --------------------------------------------------------------------------------------------------------------------- 0.5 2 3.5 5 6.5 8 + 1.5 --------------------------------------------------------------------------------------------------------------------- 7 4 1 -2 -5 -8 -3 This is a really famous number sequence which was discovered by an Italian mathematician a long time ago. It is called the Fibonacci sequence and can be seen in many natural things like pine cones and sunflowers!!! 0 1 1 2 3 5 8 13 21 etc… Can you see how it is made? What will the next number be? 34! Guess my rule! For these sequences I have done 2 maths functions! 3 7 15 31 2x +1 2 3 63 127 2x -1 5 9 17 33 What is a Number Sequence? A list of numbers where there is a pattern is called a number sequence The numbers in the sequence are said to be its members or its terms. Sequences To write the terms of a sequence given the nth term Given the expression: 2n + 3, write the first 5 terms In this expression the letter n represents the term number. So, if we substitute the term number for the letter n we will find value that particular term. The first 5 terms of the sequence will be using values for n of: 1, 2, 3, 4 and 5 term 5 term 4 term 1 term 3 term 2 2x5+3 2x1+3 2x2+3 2x3+3 2x4+3 13 5 9 11 7 Sequences Now try these: Write the first 3 terms of these sequences: 1) n+2 2) 2n + 5 7, 9, 11 3) 3n - 2 1, 4, 7 4) 5n + 3 8, 13, 18 5) -4n 6) 3, 4, 5 + 10 6, 2, - 2, n2 + 2 3, 6, 11, 6B - The General Term of A Number Sequence Sequences may be defined in one of the following ways: • listing the first few terms and assuming the pattern represented continues indefinitely • giving a description in words • using a formula which represents the general term or nth term. The first row has three bricks, the second row has four bricks, and the third row has five bricks. • If un represents the number of bricks in row n (from the top) then u1 = 3, u2 = 4, u3 = 5, u4 = 6, .... This sequence can be describe in one of four ways: • Listing the terms: • u1 = 3, u2 = 4, u3 = 5, u4 = 6, .... This sequence can be describe in one of four ways: • Using Words: The first row has three bricks and each successive row under the row has one more brick... This sequence can be describe in one of four ways: • Using an explicit formula: un = n + 2 u1 = 1 + 2 = 3 u2 = 2 + 2 = 4 u3 = 3 + 2 = 5 u4 = 4 + 2 = 6, .... This sequence can be describe in one of four ways: • Using a graph What you really need to know! An arithmetic sequence is a sequence in which the difference between any two consecutive terms, called the common difference, is the same. In the sequence 2, 9, 16, 23, 30, . . . , the common difference is 7. What you really need to know! A geometric sequence is a sequence in which the quotient of any two consecutive terms, called the common ratio, is the same. In the sequence 1, 4, 16, 64, 256, . . , the common ratio is 4. Example 1: State whether the sequence -5, -1, 3, 7, 11, … is arithmetic. If it is, state the common difference and write the next three terms. Example 2: -5, -1, 3, 7, 11, 15, 19, 23 Subtract Common difference 11 – 7 7–3 3 – -1 -1 – -5 4 4 4 4 Arithmetic! + 4 Example 2: State whether the sequence 0, 2, 6, 12, 20, … is arithmetic. If it is, state the common difference and write the next three terms. Example 2: 0, 2, 6, 12, 20 … Subtract Common difference 20 – 12 12 – 6 6–2 2–0 8 6 4 2 Not Arithmetic! Example 3: State whether the sequence 2, 4, 4, 8, 8, 16, 16 … is geometric. If it is, state the common ratio and write the next three terms. Example 3: 2, 4, 4, 8, 8, 16, 16, … Divide Common ratio 16 ÷ 16 16 ÷ 8 8÷8 8÷4 4÷4 4÷2 1 2 1 2 1 2 Not Geometric! Example 4: State whether the sequence 27, -9, 3, -1, 1/3, … is geometric. If it is, state the common ratio and write the next three terms. Example 4: 27, -9, 3, -1, 1/3, -1/9, 1/27, -1/81 Divide Common ratio 1/3 ÷ -1 -1/3 -1 ÷ 3 -1/3 3 ÷ -9 -1/3 -9 ÷ 27 -1/3 Geometric! • -1/3 Classwork Page 154 (6B) All 5. Homework Compare Arithmetic and Geometric Sequences An Arithmetic Sequence is defined as a sequence in which there is a common difference between consecutive terms. Which of the following sequences are arithmetic? Identify the common difference. 3, 1, 1, 3, 5, 7, 9, . . . YES 15.5, 14, 12.5, 11, 9.5, 8, . . . d 2 YES d 1.5 84, 80, 74, 66, 56, 44, . . . NO 8, 6, 4, 2, 0, . . . NO 50, 44, 38, 32, 26, . . . YES d 6 26, 21, 16, 11, 6, . . . The general form of an ARITHMETIC sequence. u1 First Term: Second Term: u2 u1 1d Third Term: u3 u1 2d Fourth Term: u4 u1 3d Fifth Term: u5 u1 4d nth Term: an a1 n 1 d Formula for the nth term of an ARITHMETIC sequence. un u1 n 1 d un The n term th u1 The 1st term n The term number d The common difference Given: 79, 75, 71, 67, 63, . . . Find: u32 IDENTIFY u1 79 d 4 n 32 SOLVE un u1 (n 1)d u32 79 (32 1)(4) u32 45 Given: 79, 75, 71, 67, 63, . . . Find: What term number is (-169)? IDENTIFY u1 79 d 4 un 169 SOLVE un u1 (n 1)d 169 79 (n 1)( 4) n 63 If it’s not an integer, it’s not a term in the sequence u10 3.25 Given: u12 4.25 Find: u1 What’s the real question? IDENTIFY u1 3.25 u3 4.25 n3 The Difference SOLVE un u1 n 1 d 4.25 3.25 3 1 d d 0.5 u10 3.25 Given: u12 4.25 IDENTIFY u10 3.25 d 0.5 n 10 Find: u1 SOLVE un u1 n 1 d 3.25 u1 10 1 0.5 u1 1.25 Homework Page 156 2 - 11 ( Any 8 Problems) Take Home Test Due Tuesday. Geometric Sequence • The ratio of a term to it’s previous term is constant. • This means you multiply by the same number to get each term. • This number that you multiply by is called the common ratio (r). Example: Decide whether each sequence is geometric. • 4,-8,16,-32,… • -8/4=-2 • 16/-8=-2 • -32/16=-2 • Geometric (common ratio is -2) • 3,9,-27,-81,243,… • 9/3=3 • -27/9=-3 • -81/-27=3 • 243/-81=-3 • Not geometric Rule for a Geometric Sequence un=u1r n-1 Example: Write a rule for the nth term of the sequence 5, 2, 0.8, 0.32,… . Then find u8. •First, find r. •r= 2/ 5 = .4 •un=5(.4)n-1 u8=5(.4)8-1 u8=5(.4)7 u8=5(.0016384) u8=.008192 One term of a geometric sequence is u4 = 3. The common ratio is r = 3. Write a rule for the nth term. Then graph the sequence. • If u4=3, then when n=4, un=3. • Use un=u1rn-1 • To graph, graph the points of the 3=u1(3)4-1 form (n,u ). 3 n 3=u1(3) 1/ ), • Such as, (1, 9 3=u1(27) 1/ ), (3,1), (2, 3 1/ =a 9 1 (4,3),… • un=u1rn-1 un=(1/9)(3)n-1 Two terms of a geometric sequence are u2= -4 and u6= -1024. Write a rule for the nth term. • Write 2 equations, one for each given term. u2 = u1r2-1 OR -4 = u1r u6 = u1r6-1 OR -1024 = u1r5 • Use these equations & sub in to solve for u1 & r. -4/ =u r 1 If r = 4, then u = -1. -4 5 1 -1024=( /r)r n-1 4 u =(-1)(4) -1024 = -4r n If r = -4, then u1 = 1. 256 = r4 un=(1)(-4)n-1 Both 4 = r & -4 = r Work! un=(-4)n-1 6D1 (4 a and b) • 5, 10, 20, 40 10 20 40 2 5 10 20 • So, geometric sequence with u1 = 5 r = 2 un u1r n 1 un 5 x 2 n 1 u15 5 x 2 81,920 14 6D1 (9a) • u4 = 24 u7 = 192 u1 u1 xr 24 3 u1 u1 xr 192 6 r 8 3 u1 2 24 6 u1r 192 1 3 ur 24 r2 3 u1 8 24 u1 3 un 3 x 2 n 1 Homework • Page 160 (6D.1 All) • Take Home Test Due Tuesday Compound Interest Compound Interest - Future Value Interest Interest Interest Amount $1000 133.1 1331 1210 1100 1000 100 Compounding Period 0 8 - 47 Interest 110 100 Compounding Period 1 121 121 110 100 110 100 Compounding Period 2 Time(Years) Compounding Period 3 4 COMPOUND INTEREST FORMULA r F V P V 1 n nt Where FV is the Future Value in t years and PV is the Present Value amount started with at an annual interest rate r compounded n times per year. EXAMPLE Find the amount that results from the investment: $50 invested at 6% compounded monthly after a period of 3 years. .0 6 F V 5 0 1 12 12 ( 3 ) $59.83 COMPARING COMPOUNDING PERIODS Investing $1,000 at a rate of 10% compounded annually, quarterly, monthly, and daily will yield the following amounts after 1 year: FV = PV(1 + r) = 1,000(1 + .1) = $1100.00 .1 F V 1 0 0 0 1 4 4 $ 1 1 0 3 .8 1 COMPARING COMPOUNDING PERIODS Investing $1,000 at a rate of 10% compounded annually, quarterly, monthly, and daily will yield the following amounts after 1 year: .1 F V 1 0 0 0 1 12 12 .1 F V 1 0 0 0 1 3 6 5 $ 1 1 0 4 .7 1 365 $ 1 1 0 5 .1 6 Interest Earned Investing $1,000 at a rate of 10% compounded annually, quarterly, monthly, and daily will yield the following amounts after 1 year: In t E a rn e d $ 1 1 0 4 . 7 1 - $ 1 0 0 0 $ 1 0 4 . 7 1 I . E . $ 1 1 0 5 . 1 6- $ 1 0 0 0 $ 1 0 5 . 1 6 Page 165 6D.3, #1 a & b n un 1 u1 (r ) u1 3000 r 1.1 n3 u4 3000(1.1) 3993 3 The investment will amount to $3993 B.) Interest = amount after 3 yrs – initial amount $3993 - $3000 = $993 Page 165 6D.3, #2 a & b n un 1 u1 (r ) u1 20,000 r 1.12 n4 u5 20000(1.12) 31470.39 4 The investment will amount to €31470.39 B.) Interest = €31470.39 – €31470.39 €11470.39 Homework Page 165. 6D.3 (3 – 10). 6E – Sigma Notation Vocabulary maximum value of n starting value of n k a n expression for general term n 1 Read: “the summation from n = 1 to k of an” Introduction to Sigma Notation Is read as “the sum from n equals 1 to 5 of 3n.” upper limit of summation How many terms given? 5 ∑ 3n 5 3 + 6 + 9 + 12 + 15 = ∑ 3n n=1 n=1 index of summation lower limit of summation 58 Formulas Arithmetic Sum: S n a1 an 2 a1 First Term an Last Term n Amount of Terms 59 Formulas Sigma Form of Arithmetic Series: k a n 1 1 d n 1 k Amount of Terms a1 First Term 5/7/2017 6:01 AM d Common Difference n Variable 60 Example 1 Write in Sigma Notation, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 k a n 1 1 d n 1 k 11 Amount a1 dof nTerms 1 11 11 k a1 1First 1 Term 2 n 1 1 2 1 2n Difference d Common 2 n 11 n n 2n 1 n Variable d n 1 n 1 a22n 1 61 Example 2 Write in Sigma Notation, 26 + 23 + 20 + 17 + 14 + 11 + 8 + 5 8 3n 29 5/7/2017 6:01 AM n 1 62 Your Turn Write in Sigma Notation, 4, 15, 26, …, 301 28 11n 7 n 1 5/7/2017 6:01 AM 63 Example 3 Find the following sum: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 S S n a1 an 2 111 21 2 121 64 Example 4 Find the following sum: 4, 15, 26, …, 301 5/7/2017 6:01 AM 4270 65 Your Turn Find the following sum: 15, 11, 7, …, –61 5/7/2017 6:01 AM 460 66 Example 5 32 Evaluate 32 2n 8 n 1 n a1 an 2n 8 S 2 k / n 32 2 1 8 a1 10 n 1 5/7/2017 6:01 AM an 72 2 32 8 67 Example 5 32 Evaluate 2n 8 n 1 S S 5/7/2017 6:01 AM n a1 an n 32 a1 10 an 72 2 32 10 72 2 1312 68 Example 6 4 Evaluate 5/7/2017 6:01 AM 4 n 5 n 1 20 69 Your Turn 31 Evaluate 2n 298 n 1 10, 230 5/7/2017 6:01 AM 70 Write the expression in expanded form and then find the sum. 4 2n 7 n 1 = -5 + -3 + -1 + 1 = -8 Write the expression in expanded form and then find the sum. 3 7 b 1 b4 = 80 + 242 + 728 + 2186 = 3236 Consider the Sequence 1, 4,9,16, 25,... a) Write down an expression for Sn. Sn 1 2 3 4 ... n 2 2 2 2 n k k 1 2 2 Consider the Sequence 1, 4,9,16, 25,... b) Find Sn for n = 1, 2, 3, 4, and 5 S1 1 S2 1 4 5 S3 1 4 9 14 n k k 1 S4 1 4 9 16 30 S5 1 4 9 16 25 55 2 Modeling Growth • Is this an arithmetic series or geometric series? • What is the common ration of the geometric series? u0 4000 un un 1 0.08un 1 (1 0.08)un 1 1.08un 1 Sum of a Finite Geometric Series • The sum of the first n terms of a geometric series is a1 (1 r ) Sn 1r n Notice – no last term needed!!!! Formula for the Sum of a Finite Geometric Series n What is n? What is a1? What is r? 1 r S n a1 1 r n = # of terms a1 = 1st term r = common ratio Example • Find the sum of the 1st 10 terms of the geometric sequence: 2 ,-6, 18, -54 a1 (1 r ) Sn 1r n 10 What is n? What is a1? What is r? 10 2(1 - (-3) ) 2(1 - 3 ) S10 = = 29,524 1 - -3 4 That’s It! Example: Consider the geometric series 4+2+1+½+… . • Find the sum of the first 10 terms. • Find n such that Sn=31/4. n 1 r n S n a1 1 1 1 r 31 2 10 4 1 4 1 1 1 2 2 S10 4 1 1 2 1 1023 1 2046 1023 1024 1024 4 4 S10 4 1 1 1024 128 2 2 1 n 1 31 2 4 1 4 1 2 31 1 1 32 2 32 2 n n 1 n 1 31 2 4 1 4 2 1 1 32 2 n log232=n n 31 1 8 1 2 4 1 1 32 2 n5 n 1 1n n 32 2 Assignment • When the famous mathematician C. F. Gauss was 7 years old, his teacher posed problem to the class and expected that it would keep the students busy for a long time. – Gauss, though, answered it almost immediately. • Suppose we want to find the sum of the numbers 1, 2, 3, 4, . . . , 100, that is, • 1 + 2 + 3 + 4 + 5+ 6+ …+ 100 • His idea was this: • Since we are adding numbers produced according to a fixed pattern, there must also be a pattern (or formula) for finding the sum. – He started by writing the numbers from 1 to 100 and below them the same numbers in reverse order. • Writing S for the sum and adding corresponding terms gives: S 1 2 S 100 99 3 98 99 100 98 3 2 1 2S 101 101 101 101 101 101 – It follows that 2S = 100(101) = 10,100 and so S = 5050. • We want to find the sum of the first n terms of the arithmetic sequence whose terms are un = a1 + (n – 1)d. – That is, we want to find: n Sn a1 n 1 d k 1 a1 a1 d a1 2d a1 3d a1 n 1 d • Using Gauss’s method, we write: Sn a1 a1 d a1 n 2 d a1 n 1 d Sn a1 n 1 d a1 n 2 d a1 d a1 2Sn 2a1 n 1 d 2a1 n 1 d 2a1 n 1 d 2a1 n 1 d – There are n identical terms on the right side of this equation. 2Sn n 2a1 (n 1)d n Sn 2a1 (n 1)d 2 a1 a1 (n 1)d S S n a1 a1 n 1 d 2 n 2a1 n 1 d 2 Substitute an a1 n 1 d n # of Terms n a1 an S 2 a1 1st Term d Difference • Find the sum of the first 40 terms of the arithmetic sequence 3, 7, 11, 15, . . . – Here, a = 3 and d = 4. – Using Formula 1 for the partial sum of an arithmetic sequence, we get: – S40 = (40/2) [2(3) + (40 – 1)4] = 20(6 + 156) = 3240 50 73 2 p 71 69 67 . . . 25 27 p 1 71 69 67 . . . 25 27 27 25 . . . 67 69 71 44 44 44 . . . 44 44 44 50 Terms 50 71 27 2 1100 71 + (-27) Each sum is the same. S Find the sum of the terms of this arithmetic series. 35 29 3k n a1 an k 1 2 n 35 a1 26 a35 76 35 26 76 S 2 S 875 Find the sum of the terms of this arithmetic series. 151 147 143 139 . . . 5 n a1 an S 2 n 40 a1 151 a40 5 What term is -5? an a1 n 1 d 5 151 n 1 4 n 40 40 151 5 S 2 S 2920 35 45 5i n a1 an S 2 n 35 a1 40 an 130 35 40 130 S 2 S 1575 i 1 S n 2a1 n 1 d 2 n 35 a1 40 d 5 S 35 2 40 35 1 3 S 1575 2 Homework 6F Page 169 (1 – 11) For #8, Chose 9 Just do a) & b). Formula for the Sum of a Finite Geometric Series 1 r S n a1 1 r n n = # of terms a1 = 1st term r = common ratio Example: Consider the geometric series 4+2+1+½+… . • Find the sum of the first 10 terms. • Find n such that Sn=31/4. n 1 r n S n a1 1 1 1 r 31 2 10 4 1 4 1 1 1 2 2 S10 4 1 1 2 1 1023 1 2046 1023 1024 1024 4 4 S10 4 1 1 1024 128 2 2 1 n 1 31 2 4 1 4 1 2 31 1 1 32 2 32 2 n n 1 n 1 31 2 4 1 4 2 1 1 32 2 n log232=n n 31 1 8 1 2 4 1 1 32 2 n5 n 1 1n n 32 2 Homework 6G.1 Page 171 (1 – 5) #2a (Conjugate denominator) Infinite Geometric Series • Consider the infinite geometric sequence n 1 1 1 1 1 , , , , ... ... 2 4 8 16 2 • • What happens to each term in the series? • They get smaller and smaller, but how small does a term actually get? Each term approaches 0 Partial Sums • Look at the sequence of partial sums: 1 S1 2 1 1 3 S2 2 4 4 1 1 1 7 S3 2 4 8 8 What is happening to the sum? It is approaching 1 1 0 It’s CONVERGING TO 1. Here’s the Rule Sum of an Infinite Geometric Series If |r| < 1, the infinite geometric series a 1 + a 1r + a 1r 2 + … + a 1r n + … converges to the sum a1 S 1r If |r| > 1, then the series diverges (does not have a sum) Converging – Has a Sum • So, if -1 < r < 1, then the series will converge. Look at the series given by • The graph confirms: 1 1 1 1 ... 4 16 64 256 1 4, • Since r = we know that the sum 1 a1 1 is 4 S 1r 1 1 4 3 Diverging – Has NO Sum • If r > 1, the series will diverge. Look at 1 + 2 + 4 + 8 + …. • Since r = 2, we know that the series grows without bound and has no sum. S1 1 S2 1 2 3 S3 1 2 4 7... • The graph confirms: Example • Find the sum of the infinite geometric series 9 – 6 + 4 - … 2 • We know: a1 = 9 and r = ? 3 a1 9 27 S 1 r 1 2 5 3 You Try • Find the sum of the infinite geometric series 24 – 12 + 6 – 3 + … • Since r = -½ a1 S 1 r 24 24 48 S 16 1 3 3 1 2 2 Page 173 6G.2 1. 0.3 3 3 3 .... 10 100 1000 a.) Find i. u1 ii. r 3 a.) i. u1 10 3 1 100 a.) ii. r 0.10 3 10 10 Page 173 6G.2 1. 0.3 3 3 3 .... 10 100 1000 b.) Using a, show that 0.3 1 3 b.) Sn 3 u1 n , then S 10 1 r 1 1 10 3 3 3 0.3 ... 10 100 1000 3 3 3 0.3 ... 10 100 1000 1 0.3 3 Homework Page 173 All 2 - 8 REVIEW 6A (NO CALCULATOR) REVIEW 6B (WITH CALCULATOR)
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