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UNIT 7 Sequences & Series PATTERNS EXAMPLE: o o o Suppose every student in class has a phone conversation with every other member of the class. What is the minimum number of calls required? We can show telephone conversations by using the following diagrams: ? CONTINUED…. o o o o How many calls are necessary for two people to have a conversation? 1 call How many calls are necessary for everyone to talk to everyone else in a group of three people? 3 calls In a group of four people? 6 calls Use a diagram to find the number of calls needed for five people. 10 calls CONTINUED… o Which of the following formulas can you use to find the pattern for the telephone calls? C n n 1 a.) o 2n 3 b.) n n 1 5 c.) 2 Use the formula from the previous question to find the number of calls needed for a group of 7 students. n n 1 7 7 1 7 6 42 21 2 2 2 2 o How many calls would be needed for this class? Some patterns can be described with a sequence, or a list of numbers. Each number in a sequence is a term. EXAMPLE: Describe the pattern formed & find the next three terms. a.) 27, 34, 41, 48 Add 7; 55, 62, 69 b.) 243, 81, 27, 9 Divide by 3; 3, 1, 1/3 EXAMPLE: Suppose you drop a handball from a height of 10 ft. After the ball hits the floor, it rebounds to 85% of its previous height. a.) How high will the ball rebound after its fourth bounce? Original height of the ball: 10 ft. After 1st bounce: 85% of 10 = .85 10 8.5 ft. After 2nd bounce: 85% of 8.5 = .85 8.5 7.225 After 3rd bounce: 85% of 7.225 = .85 7.225 6.141 After 4th bounce: 85% of 6.141 = .85 6.141 5.220 The ball will rebound about 5.2 feet after the 4th bounce. CONTINUED… b.) About how high will the ball rebound after the 7th bounce? After 4th bounce: 5.220 After 5th bounce: 85% of 5.220 .85 5.220 4.437 After 6th bounce: 85% of 4.437 .85 4.437 3.77145 After 7th bounce: 85% of 3.77145 .85 3.77145 3.206 The ball will rebound about 3.2 feet after the 7th bounce. CONTINUED… c.) After what bounce will the rebound height be less than 2 feet? After 7th bounce: 3.206 After 8th bounce: 85% of 3.206 .85 3.206 2.7251 After 9th bounce: 85% of 2.7251 .85 2.7251 2.3163 After 10th bounce:85% of 2.3163 .85 2.3163 1.969 After the 10th bounce, the rebound height will be less than 2 feet. HOMEWORK Worksheet: Patterns RECURSIVE FORMULA TERMS OF A SEQUENCE A variable, such as a, with positive integer subscripts, can be used to represent the terms of a sequence. 1st Term a1 2nd Term n-1 Term nth Term a2 an-1 an (previous term) n+1 Term an+1 (next term) RECURSIVE FORMULA A recursive formula defines the terms in a sequence by relating each term to the ones before it. The pattern in yesterday’s bouncing ball example was recursive because the height of the ball after each bounce was 85% of its previous height. The recursive formula that describes the ball’s height is an 0.85an1 , where a1=10 (the first term in the sequence always needs to be defined for a recursive sequence) EXAMPLE: a.) Describe the pattern that allows you to find the next term in the sequence 2, 4, 6, 8, 10…. Write a recursive formula for the sequence. Add 2 to a term to find the next term • A recursive formula is an an 1 2, where a1= 2 • b.) Find the sixth & seventh terms in the sequence. Since a5 = 10, a6 = 10+2 = 12 • Since a6 = 12, a7 = 12+2 = 14 • CONTINUED… c.) Find the value of term a9 in the sequence. Term a9 is the ninth term • a9 = a8 2 a7 2 2 14 2 2 18 • d.) Find terms a11 & a15 • a11 = a10 2 a9 2 2 18 2 2 22 • a 2 2 2 a 2 2 2 2 22 2 2 2 2 a15 = a14 2 a 2 2 13 12 11 30 EXAMPLE: Write a recursive formula for each sequence. Then find the next term. 1 1 1 1 1 , ,... a.) , , , 2 4 8 16 32 b.) 144, 36, 9, 9 ,... 4 Recursive Formula: an 1 1 an 1 , a1 2 2 Recursive Formula: an an 1 , a 144 1 4 Next Term: 1 64 Next Term: 9 16 HOMEWORK Worksheet: Recursive Formulas EXPLICIT FORMULAS EXPLICIT FORMULA A recursive formula allows us to find the value of a term by using the preceding term. However, we can sometimes find the value of a term without knowing the preceding term by using the number of the term to calculate its value. An explicit formula expresses the nth term in terms of n. EXAMPLE: The spreadsheet shows the perimeters of squares with sides from 1 to 6 units long. The numbers in each row form a sequence. a.) For each sequence, find the next term (a7) & the twentyfifth term (a25). Row 2: each term is the same as its subscript; therefore, a7 = 7 and a25 = 25 Row 3: each term is 4 times its subscript; therefore, a7 = 4(7) = 28 & a25 = 4(25) = 100. CONTINUED… b.) Write an explicit formula for each sequence. Row 2: Row 3: an n an 4n c.) Write the first six terms in the sequence showing the areas of the squares. Then find a20 Area: 1, 4, 9, 16, 25, 36; 400 d.) Write an explicit formula for the sequence from part (c). a n2 n HOMEWORK Worksheet: Explicit Formulas ARITHMETIC SEQUENCES ARITHMETIC SEQUENCES In an arithmetic sequence, the difference between consecutive terms is constant. This difference is called the common difference. The common difference can be positive (the terms of the sequence are increasing in value) or negative (the terms of the sequence are decreasing in value). EXAMPLE: Is the given sequence arithmetic? If so, identify the common difference. a.) 2, 4, 8, 16,… There is no common difference. It is not an arithmetic sequence. +2 +4 +8 b.) 6, 12, 18, 24,… +6 +6 +6 There is a common difference, therefore this is an arithmetic sequence. The common difference is 6. You can use an explicit formula to find the value of the nth term of an arithmetic sequence when the previous term is unknown. ARITHMETIC SEQUENCE FORMULAS Recursive Formula: a1 a given value, an an1 d Explicit Formula: an a1 n 1 d In these formulas, an is the nth term, a1 is the 1st term, n is the number of the term, and d is the common difference. EXAMPLE: Suppose you participate in a bike-a-thon for charity. • The charity starts with $1100 in donations. • Each participant must raise at least $35 in pledges. • What is the minimum amount of money raised if there are 75 participants? Why the 76th Find the 76th term of the sequence, 1100, 1135, 1170,… • Use the explicit formula Substitute a1 = 1100, n =7 6, & d = 35 an a1 n 1 d a76 1100 76 1 35 a76 1100 75 35 a76 1100 2625 a76 3725 term? With 75 participants, the bike-a-thon will raise a minimum of $3,725 EXAMPLE: • Use the explicit formula to find the 25th term on the sequence 5, 11, 17, 23, 29 Use the explicit formula Substitute a1 = 5, n =25, &d=6 an a1 n 1 d a25 5 25 1 6 a25 5 24 6 a25 5 144 a25 149 ARITHMETIC MEAN The arithmetic mean of any two numbers is the average of the two numbers sum of two numbers Arithmetic Mean = 2 For any three sequential terms in an arithmetic sequence, the middle term is the arithmetic mean of the first and third terms. You can use arithmetic mean to find a missing term of an arithmetic sequence. EXAMPLE Find the missing term of each arithmetic sequence. a.) 84, ? , 110 Arithmetic Mean = sum of two numbers 2 84 110 = 2 194 = 2 = 97 b.) 24, ? , 57 Arithmetic Mean = sum of two numbers 2 24 57 = 2 81 = 2 = 40.5 HOMEWORK Worksheet: Arithmetic Sequences GEOMETRIC SEQUENCES GEOMETRIC SEQUENCES The sequence 1, 2, 4, 8,… is a geometric sequence. How would you describe the sequence? Multiply by 2 The sequence 900, 300, 100,… is also a geometric sequence. How would you describe the sequence? Divide by 3 or Multiply by 1/3 GEOMETRIC SEQUENCES What is common between geometric sequences? In geometric sequences, the ratio between consecutive terms is constant. This ratio is called the common ratio. EXAMPLE: Is the given sequence geometric? If so, identify the common ratio. a.) 5, 15, 45, 135,… Each term is being multiplied by 3. So, yes, it is a geometric sequence. x3 x3 x3 b.) 15, 30, 45, 60,… x 2 x 1.5 x 1.3 The common ratio is 3 No common ratio; therefore, not a geometric sequence. GEOMETRIC SEQUENCE FORMULAS Recursive Formula: a1 a given value, an an1 r n 1 a a r Explicit Formula: n 1 In these formulas, an is the nth term, a1 is the 1st term, n is the number of the term, and r is the common ratio. EXAMPLE: • • • • Suppose you want a reduced copy of a photograph. The actual length of the photograph is 10 in. The smallest size the copier can make is 64% of the original. Find the length of the photograph after five reductions of 64%. For five reductions, you need to find the 6th term of the geometric sequence 10, 6.4… Use the explicit formula Substitute a1 = 10, n = 6, & r = 0.64 an a1 r n 1 a6 10 0.6461 a6 10 0.645 a6 10 0.1074 a6 1.07 After five reductions of 64%, the photograph is about 1 in. long GEOMETRIC MEAN You can find the geometric mean of any two positive numbers by taking the positive square root of the product of the two numbers. Geometric Mean = product of two #'s You can use the geometric mean to find a missing term of a geometric sequence. EXAMPLE Find the missing term of each geometric sequence. a.) 12, ?, 3 G.M. = product of two #'s b.) 5, ? , 2.8125 G.M. = product of two #'s = 12 3 = 5 2.8125 = 36 = 14.0625 =6 = 3.75 HOMEWORK Worksheet: Geometric Sequences ARITHMETIC SERIES – DAY 1 SERIES • A series is the expression for the sum of the terms of a sequence Finite sequences and series have terms that you can count individually from 1 to a whole number n. • Infinite sequences and series continue without end. • Finite Sequence Finite Series 6, 9, 12, 15, 18 6 + 9 + 12 + 15 + 18 Infinite Sequence 3, 7, 11, 15, … Infinite Series 3 + 7 + 11 + 15 + … EXAMPLE: Write the related series for each finite sequence. Then evaluate the series. a.) 2, 11, 20, 29, 38, 47 Related Series: 2 + 11 + 20 + 29 + 38 + 47 To evaluate, add the series The sum of the terms of the sequence = 147 b.) 0.3, 0.6, 0.9, 1.2, 1.5, 1.8, 2.1, 2.4, 2.7, 3.0 Related Series: 0.3 + 0.6 + 0.9 + 1.2 + 1.5 + 1.8 + 2.1 + 2.4 + 2.7 + 3.0 The sum of the terms of the sequence = 16.5 c.) 100, 125, 150, 175, 200, 225 Related Series: 100 + 125 + 150 + 175 + 200 + 225 The sum of the terms of the sequence = 975 ARITHMETIC SERIES • An arithmetic series is a series of terms whose terms form an arithmetic sequence. • When a sequence has many terms, or when you know only the first & last terms of the sequence, you can use a formula to evaluate the related series quickly. SUM OF A FINITE ARITHMETIC SERIES The sum Sn of a Finite Arithmetic Series a1 + a2 + a3 + … + an is n Sn a1 an 2 where a1 is the first term, an is the nth term, and n is the number of terms. EXAMPLE: xxxxx xxxxxxx • Several rows of cross-stitches make up the xxxxxxxxx green roof in the picture to the right. xxxxxxxxxxx a.) Find the total number of green cross-stitches xxxxxxxxxxxxx in the roof. The # of cross-stitches in each row form a series. Row 1 = 5 stitches, Row 2 = 7 stitches, … Sum of a finite arithmetic series Substitute n = 5, a1 = 5, & an = 13 n Sn a1 an 2 5 5 S n 5 13 18 45 2 2 There are 45 stitches in the green roof. HOMEWORK Worksheet: Arithmetic Series – Day 1 ARITHMETIC SERIES – DAY 2 SUMMATION NOTATION You can use the summation symbol ∑ to write a series. Then you can use limits to indicate how many terms you are adding. Limits are the least and greatest integral values of n. upper limit, greatest value of n lower limit, least value of n 3 5n 1 n 1 explicit formula for the sequence EXAMPLE: Use summation notation to write each series for the specified number of terms. a.) 3 + 6 + 9 + … for 33 terms Write the explicit formula Substitute a1 = 3 & d = 3 n = the # of the term, so leave that blank distribute & simplify upper limit = 33 (greatest value of n) lower limit (least value of n) =1 an a1 n 1 d an 3 n 1 3 an 3 3n 3 an 3n 33 3n n 1 CONTINUED… b.) 1 + 2 + 3 + … ; n = 6 Write the explicit formula an a1 n 1 d Substitute a1 = 1 & d = 1 an 1 n 11 n = the # of the term, so leave that blank distribute & simplify upper limit =6 (greatest value of n) lower limit (least value of n) =1 an 1 n 1 an n 6 n n 1 CONTINUED… c.) 3 + 8 + 13 + 18 … ; n = 9 Write the explicit formula an a1 n 1 d Substitute a1 = 3 & d = 5 an 3 n 1 5 n = the # of the term, so leave that blank distribute & simplify upper limit =9 (greatest value of n) lower limit (least value of n) =1 an 3 5n 5 an 5n 2 9 5n 2 n 1 EXAMPLE: For each sum, find the number of terms, the first term, and the last term. Then evaluate the series. 3 a.) 5n 1 n 1 Step 1: Find the number of terms Since the values of n are 1, 2, & 3, there are 3 terms in the series Step 2: Find the 1st & last terms 5n 1 5 1 1 6 last term: 5n 1 5 3 1 16 1st term: Step 3: Evaluate Use formula n Sn a1 an 2 n 3 3 Sn a1 an 6 16 22 33 2 2 2 CONTINUED… 10 b.) n 3 n 1 Step 1: Find the number of terms Since the values of n are 1 thru 10, there are 10 terms in the series Step 2: Find the 1st & last terms 1st term: last term: n 3 1 3 2 n 3 10 3 7 n Step 3: Evaluate Use formula S n a1 an 2 n 10 Sn a1 an 2 7 5 5 25 2 2 CONTINUED… 5 c.) 2 n n2 Step 1: Find the number of terms Since the values of n are 2 thru 5, there are 4 terms in the series Step 2: Find the 1st & last terms 1st term: n 2 22 4 last term: n 2 52 25 Step 3: Evaluate Use formula n Sn a1 an 2 4 n Sn a1 an 4 25 2 29 58 2 2 HOMEWORK Worksheet: Arithmetic Series - Day 2 GEOMETRIC SERIES – DAY 1 Evaluating a Finite Geometric Series GEOMETRIC SERIES A geometric series is the expressions for the sum of the terms of a geometric sequence. SUM OF A FINITE GEOMETRIC SERIES The sum Sn of a Finite Geometric Series a1 + a2 + a3 + … + an, r ≠ 1, is Sn a1 1 r n 1 r where a1 is the first term, r is the common ratio, and n is the number of terms. EXAMPLE: Evaluate the series to the given term. a.) 3 + 6 + 12 + 24…; S6 Use the formula Substitute a1 = 3, r = 2, & n = 6 Simplify Sn S6 a1 1 r n 1 r 3 1 26 1 2 3 1 64 3 63 189 1 1 1 189 CONTINUED… b.) -45 + 135 - 405 +…; S5 Use the formula Substitute a1 = -45, r = -3, & n = 5 Simplify Sn S5 a1 1 r n 1 r 45 1 (3)5 1 (3) 45 1 (243) 45 244 4 4 10980 2745 4 EXAMPLE: • • • In March, the Floyd family starts saving for a vacation in August, which they expect to cost $1375. They start with $125 and each month they plan to deposit 20% more than the previous month. Will they have enough money for their trip? Use the formula Substitute a1 = 125, r = 1.2, & n = 6 Simplify Sn a1 1 r n 1 r 125 1 1.26 1 1.2 125 1 2.986 125 1.986 0.2 0.2 248.25 $1,241.25 0.2 No, the Floyd family will not have enough money for their vacation. HOMEWORK Worksheet: Geometric Series - Day 1 GEOMETRIC SERIES – DAY 2 Evaluating an Infinite Geometric Series INFINITE GEOMETRIC SERIES In some cases, you can evaluate an infinite geometric series. When r 1, the series converges, or gets closer and closer, to the sum S. It will eventually have a sum. When r 1 , the series diverges, or approaches no limit. EXAMPLE: Decide whether each infinite geometric series diverges or converges. State whether the series has a sum. a.) 1 1 1 ... 3 9 a1 = 1 & a2 = -1/3 therefore, r = 1 3 Since r 1, the series converges and the series has a sum b.) 5 2 n 1 n 1 a1 = 5(2)1-1 = 5(2)0 = 5(1)= 5 a2 = 5(2)2-1 = 5(2)1 = 5(2) = 10 Since a1 = 5 and a2 = 10, r=2 Since r 1, the series diverges and the series does not have a sum CONTINUED… Decide whether each infinite geometric series diverges or converges. State whether the series has a sum. c.) 1 1 1 ... 5 25 a1 = 1 & a2 = 1/5 therefore, r = 1 5 Since r 1, the series converges and the series has a sum d.) 4 8 16 .... a1 = 4 & a2 = 8 therefore, r = 2 Since r 1, the series diverges and the series does not have a sum SUM OF AN INFINITE GEOMETRIC SERIES An infinite geometric series with r 1 converges to the sum a1 S 1 r Where a1 is the first term and r is the common ratio. EXAMPLE: Evaluate each infinite geometric series 1 1 1 a.) 1 ... 2 4 8 1 1 a1 2 S 1 r 1 1 1 2 2 3 3 3 b.) 3 ... 2 4 8 a1 S 1 r 3 3 2 1 1 1 1 2 2 EXAMPLE: • • The length of the outside shell of each closed chamber of a chambered nautilus is 0.9 times the length of the larger chamber next to it. Estimate the total length of the outside shell for the enclosed chambers. The outside edge of the largest enclosed chamber is 27 mm long, so a1 = 27 Use the formula Substitute a1 = 27 & r = 0.9 Simplify a1 S 1 r 27 1 0.9 27 0.1 270 HOMEWORK Worksheet: Geometric Series – Day 2