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Exercises 1 As the exercise mentions, a recursive rule defines a sequence by giving the first term or terms. Furthermore, it tells how the next term is related to the previous term or terms. This relationship can be written algebraically as an equation which is known as the recursive equation. An example is shown below. Example initial termsa1=1 and a2=2Recurisive Equationan=(an−1)(an−2)Sequence1,2,2,4,8,32,256… With this in mind, we can complete the exercise's sentence.A recursive rule gives the beginning term(s) of the sequence, and a recursive equation tells how an is related to one or more preceding term. | |

Exercises 2 We are given the rules shown below and we are asked to find the one that is different. a1=-1, 5an−1a1=-3, an=an−1+1an=6n−2a1=9, an=4an−1 Recall that a rule can be explicit when it gives the nth term, an, as a function of the term's position number n. Or, it is recursive if it gives the first term or terms, and a recursive equation that tells how an is related to the preceding terms. Let's take a second look at the rules given and identify which of them are recursive rules. a1=-1, 5an−1a1=-3, an=an−1+1an=6n−2a1=9, an=4an−1 As we can see, all the rules are recursive since they give a first term and a recursive equation, except for an=6n−2. This rule is explicit, as it just gives an equation for an in terms of the term's position. Therefore, this is the different one. | |

Exercises 3 We will use the following facts.The recursive equation for an arithmetic sequence is an=an−1+d. The recursive equation for a geometric sequence is an=r⋅an−1.We are given the recursive rule below. a1=2, an=7⋅an−1 Since the recursive equation is of the form an=r⋅an−1, it represents a geometric sequence. | |

Exercises 4 We will use the following facts.The recursive equation for an arithmetic sequence is an=an−1+d. The recursive equation for a geometric sequence is an=r⋅an−1.We are given the recursive rule below. a1=18, an=an−1+1 Since the recursive equation is of the form an=an−1+d, it represents an arithmetic sequence. | |

Exercises 5 We will use the following facts.The recursive equation for an arithmetic sequence is an=an−1+d. The recursive equation for a geometric sequence is an=r⋅an−1.We are given the recursive rule below. a1=5, an=an−1−4 ⇕a1=5, an=an−1+(-4) Since the recursive equation is of the form an=an−1+d, it represents an arithmetic sequence. | |

Exercises 6 We will use the following facts.The recursive equation for an arithmetic sequence is an=an−1+d. The recursive equation for a geometric sequence is an=r⋅an−1.We are given the recursive rule below. a1=3, an=-6⋅an−1 Since the recursive equation is of the form an=r⋅an−1, it represents a geometric sequence. | |

Exercises 7 We are asked to write the first 6 terms of a sequence, given a recursive rule. a1an=0=an−1+2, for n≥2 To do so, we will use a table.nan=an−1+2an−1+2an 1a1=0−0 2a2=a2−1+2a1+2=0+22 3a3=a3−1+2a2+2=2+24 4a4=a4−1+2a3+2=4+26 5a5=a5−1+2a4+2=6+28 6a6=a6−1+2a5+2=8+210 Therefore, the first 6 terms of the sequence are 0, 2, 4, 6, 8, and 10. Let's graph these points! | |

Exercises 8 We are asked to write the first 6 terms of a sequence, given a recursive rule. a1an=10=an−1−5, for n≥2 To do so, we will use a table.nan=an−1−5an−1−5an 1a1=10−10 2a2=a2−1−5a1−5=10−55 3a3=a3−1−5a2−5=5−50 4a4=a4−1−5a3−5=0−5-5 5a5=a5−1−5a4−5=-5−5-10 6a6=a6−1−5a5−5=-10−5-15 Therefore, the first 6 terms of the sequence are 10, 5, 0, -5, -10, and -15. Let's graph these points! | |

Exercises 9 We are asked to write the first 6 terms of a sequence, given a recursive rule. a1an=2=3an−1 To do so, we will use a table.nan=3an−13an−1an 1a1=2−2 2a2=3a2−13a1=3⋅26 3a3=3a3−13a2=3⋅618 4a4=3a4−13a3=3⋅1854 5a5=3a5−13a4=3⋅54162 6a6=3a6−13a5=3⋅162486 Therefore, the first 6 terms of the sequence are 2, 6, 18, 54, 162, and 486. To graph the first six terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 10 We are asked to write the first 6 terms of a sequence, given its recursive rule. a1an=8=1.5an−1, for n>1 To do so, we will use a table.nan=1.5an−11.5an−1an 1a1=8−8 2a2=1.5a2−11.5a1=1.5(8)12 3a3=1.5a3−11.5a2=1.5(12)18 4a4=1.5a4−11.5a3=1.5(18)27 5a5=1.5a5−11.5a4=1.5(27)40.5 6a6=1.5a6−12.1.5a5=1.5(40.5)60.75 The first 6 terms of the sequence are 8, 12, 18, 27, 40.5, and 60.75. To graph them, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 11 We are asked to write the first 6 terms of a sequence, given a recursive rule. a1an=80=-21an−1 To do so, we will use a table.nan=-21an−1-21an−1an 1a1=80−80 2a2=-21a2−1-21a1=-21(80)-40 3a3=-21a3−1-21a2=-21(-40)20 4a4=-21a4−1-21a3=-21(20)-10 5a5=-21a5−1-21a4=-21(-10)5 6a6=-21a6−1-21a5=-21(5)-25 Therefore, the first 6 terms of the sequence are 80, -40, 20, -10, 5, and -25. To graph the first six terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 12 We are asked to write the first 6 terms of a sequence, given a recursive rule. a1an=-7=-4an−1 To do so, we will use a table.nan=-4an−1-4an−1an 1a1=-7−-7 2a2=-4a2−1-4a1=-4⋅(-7)28 3a3=-4a3−1-4a2=-4⋅28-112 4a4=-4a4−1-4a3=-4⋅(-112)448 5a5=-4a5−1-4a4=-4⋅448-1792 6a6=-4a6−1-4a5=-4⋅(-1792)7168 Therefore, the first 6 terms of the sequence are -7, 28, -112, 448, -1792, and 7168. To graph the first six terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 13 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is arithmetic, with first term a1=7 and common difference r=9. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of a arithmetic sequence. a1=aan=an−1+d⇒a1=7an=an−1+9 | |

Exercises 14 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is geometric, with first term a1=8 and a common ratio of r=3. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of a geometric sequence. a1=aan=r⋅an−1⇒a1=8an=3an−1 | |

Exercises 15 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is geometric, with first term a1=243 and common ratio r=0.3, or 31. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of a geometric sequence. a1=aan=r⋅an−1⇒a1=243an=31an−1 | |

Exercises 16 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is arithmetic with first term a1=3 and common difference d=8. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of an arithmetic sequence. a1=aan=an−1+d⇒a1=3an=an−1+8 | |

Exercises 17 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is arithmetic, with first term a1=0 and common difference r=-3. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of a arithmetic sequence. a1=aan=an−1+d⇒a1=0an=an−1−3 | |

Exercises 18 To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern.The sequence is geometric, with first term a1=5 and common ratio r=-4. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of a geometric sequence. a1=aan=r⋅an−1⇒a1=5an=-4an−1 | |

Exercises 19 To write a recursive formula for the given sequence, we will use a table to organize the terms given in the graph and find the pattern.The sequence is geometric, with first term a1=-1 and common ratio r=4. To write the recursive formula, we will substitute these values into the general formula for the recursive rule of a geometric sequence. a1=a;an=r⋅an−1⇒a1=-1,an=4an−1 | |

Exercises 20 Let's look at the given graph. The points represent terms of the sequence.To write a recursive rule for the given sequence, we will use a table to organize the terms and find the pattern. Let's put in the table coordinates of the points from the graph.The sequence is arithmetic with first term a1=35 and common difference d=-11. To write the recursive rule, we will substitute these values into the general formula for the recursive rule of an arithmetic sequence. a1=aan=an−1+d⇒a1=35an=an−1−11 | |

Exercises 21 Let's make a table that shows the number of bacterial cells over time.From the table, we see that there is a common ratio of 2, so it is a geometric sequence. Since r=2, we can write the recursive equation for the number of bacterial cells over time. an=r⋅an−1⇒an=2⋅an−1 The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=1, an=2⋅an−1 | |

Exercises 22 Let's make a table that shows the length of the deer antler over time.From the table, we see that there is a common difference of 2, so it is an arithmetic sequence. Since d=2, we can write the recursive equation for the length of the deer antler. an=an−1+d⇒an=an−1+41 The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=421, an=an−1+41 | |

Exercises 23 The explicit formula of an arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: an=an−1+d; a1=a1 an=a1+(n−1)d In these formulas, d is the common difference and a1 is the first term. Looking at the given recursive formula, we can identify the common difference d and the value of the first term a1. an=an−1+3;a1=-3 We can see that 3 is the common difference and the first term is -3. Now we have enough information to form an explicit formula for this sequence. an=a1+(n−1)da1=-3, d=3an=-3+(n−1)3Distribute 3an=-3+3n−3Subtract termsan=3n−6 | |

Exercises 24 The explicit formula of an arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1, an=an−1+d; an=a1+(n−1)d In these formulas, d is the common difference and a1 is the first term. Looking at the given recursive formula, we can identify the common difference d and the value of the first term a1. a1a1=8;an=an−1−12⇕=8;an=an−1+(-12) We can see that -12 is the common difference and the first term is equal 8. Now we have enough information to form an explicit formula for this sequence. an=a1+(n−1)da1=8, d=-12an=8+(n−1)(-12) Simplify right-hand side Distribute -12an=8−12n+12Add terms an=-12n+20 | |

Exercises 25 The explicit formula of an geometric sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: an=r⋅an−1; a1=a1 an=a1(r)n−1 In these formulas, r is the common ratio and a1 is the first term. Looking at the given recursive formula, we can identify the common ratio r and the value of the first term a1. an=0.5an−1;a1=16 We can see that 0.5 is the common ratio and the first term is 16. Now we have enough information to form an explicit formula for this sequence. an=a1(r)n−1a1=16, r=0.5an=16(0.5)n−1 | |

Exercises 26 The explicit formula of a geometric sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1; an=r⋅an−1 an=a1⋅rn−1 In these formulas, r is the common ratio and a1 is the first term. Looking at the given recursive formula, we can identify the common ratio r and the value of the first term a1. a1=-2; an=9an−1 We can see that 9 is a common ratio and -2 is the first term. Now we have enough information to form an explicit formula for this sequence. an=a1⋅rn−1⇔an=-2(9)n−1 | |

Exercises 27 The explicit formula of an arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: an=an−1+d; a1=a1 an=a1+(n−1)d In these formulas, d is the common difference and a1 is the first term. Looking at the given recursive formula, we can identify the common difference d and the value of the first term a1. an=an−1+17;a1=4 We can see that 17 is the common difference and the first term is 4. Now we have enough information to form an explicit formula for this sequence. an=a1+(n−1)da1=4, d=17an=4+(n−1)17Distribute 3an=4+17n−17Subtract termsan=17n−13 | |

Exercises 28 The explicit formula of an geometric sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: an=r⋅an−1; a1=a1 an=a1(r)n−1 In these formulas, r is the common ratio and a1 is the first term. Looking at the given recursive formula, we can identify the common ratio r and the value of the first term a1. an=-5an−1;a1=5 We can see that -5 is the common ratio and the first term is 5. Now we have enough information to form an explicit formula for this sequence. an=a1(r)n−1a1=5, r=-5an=5(-5)n−1 | |

Exercises 29 The explicit formula of a geometric sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1; an=r⋅an−1 an=a1rn−1 In these formulas, r is the common ratio and a1 is the first term. Looking at the given explicit formula, we can identify the common ratio r and the value of the first term a1. an=7(3)n−1 We can see that 3 is the common ratio and the first term is 7. Now we have enough information to form a recursive formula for this sequence. a1=a1an=r⋅an−1⇒a1=7an=3an−1 | |

Exercises 30 The explicit formula of an arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Explicit: Recursive: an=a1+(n−1)d; a1=a1, an=an−1+d In these formulas, d is the common difference and a1 is the first term. Looking at the given explicit formula, we can identify the common difference d as the coefficient of the n-term. Then we will calculate the value of the first term a1. an=-4n+2 We can see that -4 is the common difference. We can calculate the first term by substituting n=1 into the given explicit formula. a1=-4(1)+2=-4+2=-2 Now we have enough information to form a recursive formula for this sequence. a1a1=-2, an=an−1+(-4)⇕=-2, an=an−1−4 | |

Exercises 31 The explicit formula of a arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1; an=an−1+d an=a1+(n−1)d In these formulas, d is the common difference and a1 is the first term. Looking at the given explicit formula, we can't identify the common difference d nor the value of the first term a1. So let's rewrite the formula an=1.5n+3Write as a differencean=1.5n+4.5−1.5Commutative Property of Additionan=4.5+1.5n−1.5Factor out 1.5an=4.5+1.5(n−1) Now we can identify the common difference d and the value of the first term a1. an=4.5+1.5(n−1) We can see that 1.5 is the common difference and the first term is 4.5. Now we have enough information to form a recursive formula for this sequence. a1=a1an=an−1+d⇒a1=4.5an=an−1+1.5 | |

Exercises 32 The explicit formula of a arithmetic sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1; an=an−1+d an=a1+(n−1)d In these formulas, d is the common difference and a1 is the first term. Looking at the given explicit formula, we can't identify the common difference d nor the value of the first term a1. So let's rewrite the formula an=6n−20Write as a suman=6n−6−14Factor out 6an=6(n−1)−14Commutative Property of Additionan=-14+6(n−1) Now we can identify the common difference d and the value of the first term a1. an=-14+6(n−1) We can see that 6 is the common difference and the first term is -14. Now we have enough information to form a recursive formula for this sequence. a1=a1an=an−1+d⇒a1=-14an=an−1+6 | |

Exercises 33 The explicit formula of a geometric sequence combines the information provided by the two equations of the recursive form into a single equation. Recursive: Explicit: a1=a1; an=r⋅an−1 an=a1rn−1 In these formulas, r is the common ratio and a1 is the first term. Looking at the given explicit formula, we can identify the common ratio r and the value of the first term a1. an=(-5)n−1 We can see that -5 is the common ratio and the first term is 1. Now we have enough information to form a recursive formula for this sequence. a1=a1an=r⋅an−1⇒a1=1an=-5an−1 | |

Exercises 34 We want to write the recursive formula for the given geometric sequence. Looking at the explicit formula, we can identify the common ratio r and the value of the first term a1. an=-81(32)n−1 We can see that 32 is the common ratio and the first term is -81. Now we have enough information to form an recursive formula for this sequence. a1=aan=r⋅an−1⇒a1=-81an=(32)an−1 | |

Exercises 35 We will find the first four terms then graph them. After that, we will write a recursive and an explicit rule for the sequence.Graph We will make a table to show the first four terms. Note that the first term is 5 and the common difference is 15.Let's plot the ordered pairs (1,5), (2,20), (3,35), and (4,50).Recursive Rule The sequence an is an arithmetic sequence, with first term a1=5 and common difference d=15. Then we can write the recursive equation. an=an−1+d⇒an=an−1+15 The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=5, an=an−1+15Explicit Rule The explicit rule for an arithmetic sequence is the formula an=a1+(n−1)d. For our sequence, a1=5 and d=15. an=a1+(n−1)d⇒an=5+(n−1)15 Let's simplify the right hand side of the equation. an=5+(n−1)15Distribute 15an=5+15n−15Subtract terman=15n−10 This equation is the explicit rule for the sequence. Explicit Rule: an=15n−10 | |

Exercises 36 We will find the first four terms then graph them. After that, we will write a recursive and an explicit rule for the sequence.Graph We will make a table to show the first four terms. Note that the first term is 16 and the common ratio is 21.Let's plot the ordered pairs (1,16), (2,8), (3,4), and (4,2).Recursive Rule The sequence an is a geometric sequence, with first term a1=16 and common ratio r=21. Then we can write the recursive equation. an=r⋅an−1⇒an=21⋅an−1 The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=16, an=21⋅an−1Explicit Rule The explicit rule for a geometric sequence is the formula an=a1(r)n−1, where a1 is the first term and r is the common ratio. We know a1=16 and r=21 Explicit Rule: an=16(21)n−1 | |

Exercises 37 We will find the first four terms then graph them. After that, we will write a recursive and an explicit rule for the sequence.Graph We will make a table to show the first four terms. Note that the first term is -1 and the common ratio is -3.Let's plot the ordered pairs (1,-1), (2,3), (3,-9), and (4,27).Recursive Rule The sequence an is a geometric sequence, with first term a1=-1 and common ratio r=-3. Then we can write the recursive equation. an=r⋅an−1⇒an=-3⋅an−1 The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=-1, an=-3⋅an−1Explicit Rule The explicit rule for a geometric sequence is the formula an=a1(r)n−1, where a1 is the first term and r is the common ratio. We know a1=-1 and r=-3 an=a1(r)n−1⇒an=-1(-3)n−1 Since multiplying any number by 1 does not change the result, we can remove it. Explicit Rule: an=-(-3)n−1 | |

Exercises 38 We will find the first four terms then graph them. After that, we will write a recursive and an explicit rule for the sequence.Graph We will make a table to show the first four terms. Note that the first term is 19 and the common difference is -13.Let's plot the ordered pairs (1,19), (2,6), (3,-7), and (4,-20).Recursive Rule The sequence an is an arithmetic sequence, with first term a1=19 and common difference d=-13. Then we can write the recursive equation. an=an−1+d⇒an=an−1+(-13) The recursive rule is the recursive equation together with the first term. Recursive Rule: a1=19, an=an−1−13Explicit Rule The explicit rule for an arithmetic sequence is the formula an=a1+(n−1)d. For our sequence, a1=19 and d=-13. an=a1+(n−1)d⇒an=19+(n−1)(-13) Let's simplify the right hand side of the equation. an=19+(n−1)(-13)Distribute -13an=19−13n+13Add termsan=-13n+32 This equation is the explicit rule for the sequence. Explicit Rule: an=-13n+32 | |

Exercises 39 We want to write a recursive formula for the given sequence. 1,3,4,7,11,… To do so we need to analyze how the consecutive terms are related. Let's find the sum between each pair of consecutive terms. a2+a1a3+a2a4+a3⋮===3+14+37+4⋮===4711⋮1111an−1+an−2=an1111We can see above that the sum of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to add two previous terms. With this information and knowing that the first term equals 1 and the second term equals 3, we can write the recursive formula. a1=1, a2=3 and an=an−1+an−2 We will write the next 2 terms of a sequence now. ⋮a5+a4a6+a5==⋮11+718+11==⋮1829 | |

Exercises 40 We want to write a recursive formula for the given sequence. 10,9,1,8,-7,15,… To do so we need to analyze how the consecutive terms are related. Let's find the difference between each pair of consecutive terms. a1−a2a2−a3a3−a4a4−a5⋮====10−99−11−88−(-7)⋮====18-715⋮1111an−2−an−1=an1111We can see above that the difference of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to subtract two previous terms. With this information and knowing that the first term equals 10 and the second term equals 9, we can write the recursive formula. a1=10, a2=9 and an=an−2−an−1 We will write the next 2 terms of a sequence now. ⋮a5−a6a7−a8==⋮-7−1515−(-22)==⋮-2237 | |

Exercises 41 We want to write a recursive formula for the given sequence. 2,4,2,-2,-4,-2,… To do so we need to analyze how the consecutive terms are related. Let's find the difference between each pair of consecutive terms. a2−a1a3−a2a4−a3a5−a4⋮====4−22−4-2−2-4−(-2)⋮====2-2-4-2⋮1111an−1−an−2=an1111We can see above that the difference of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to subtract two previous terms. With this information and knowing that the first term equals 2 and the second term equals 4, we can write the recursive formula. a1=2, a2=4 and an=an−1−an−2 We will write the next 2 terms of a sequence now. ⋮a6−a5a7−a6==⋮-2−(-4)2−(-2)==⋮24 | |

Exercises 42 We want to write a recursive formula for the given sequence. 6,1,7,8,15,23,… To do so we need to analyze how the consecutive terms are related. Let's find the sum between each pair of consecutive terms. a2+a1a3+a2a4+a3a5+a4⋮====1+67+18+715+8⋮====781523⋮1111an−1+an−2=an1111We can see above that the sum of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to add two previous terms. With this information and knowing that the first term equals 6 and the second term equals 1, we can write the recursive formula. a1=6, a2=1 and an=an−1+an−2 We will write the next 2 terms of a sequence now. ⋮a6+a5a7+a6==⋮23+1538+23==⋮3861 | |

Exercises 43 We want to write a recursive formula for the given sequence. 1,3,3,9,27… To do so we need to analyze how the consecutive terms are related. Let's find the product of each pair of consecutive terms. a2⋅a1a3⋅a2a4⋅a3⋮===3⋅13⋅39⋅3⋮===3927⋮1111an−1⋅an−2=an1111We can see above that the sum of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to multiply two previous terms. With this information and knowing that the first term equals 1 and the second term equals 3, we can write the recursive formula. a1=1, a2=3 and an=an−1⋅an−2 We will write the next 2 terms of a sequence now. ⋮a5⋅a4a6⋅a5==⋮27⋅9243⋅27==⋮2436561 | |

Exercises 44 We want to write a recursive formula for the given sequence. 64,16,4,4,1… To do so we need to analyze how the consecutive terms are related. Let's find the product of each pair of consecutive terms. a2a1a3a2a4a3⋮===166441644⋮===441⋮1111an−1an−2=an1111We can see above that the sum of the consecutive terms equals the next term. Therefore, to obtain the value of the term in the nth position, we need to divide two previous terms. With this information and knowing that the first term equals 1 and the second term equals 3, we can write the recursive formula. a1=64, a2=16 and an=an−1an−2 We will write the next 2 terms of a sequence now. ⋮a5a4a6a5==⋮1441==⋮441 | |

Exercises 45 Recall that the recursive equation for an arithmetic sequence is of the following form. an=an−1+d Here, d is the common difference. Then, for the given sequence, the common difference is -12. an=an−1−12⇔an=an−1+(-12) The explicit formula for the sequence, with first term a1=6 and common difference -12, can be written as follows. an=a1+(n−1)d⇒an=6+(n−1)(-12) Let's simplify it. an=6+(n−1)(-12)Distribute (-12)an=6−12n+12Add termsan=18−12n The explicit rule for the sequence is an=18−12n. Correct: Incorrect: an=18−12n✓ an=-6+12n× | |

Exercises 46 Let's check if the difference between the other consecutive terms is 2.We see that 2 is not a common difference meaning the sequence is not an arithmetic sequence. The recursive rule written in the solution is not the correct rule. Incorrect: a1=2, an=an−1+2× Let's check for a common ratio.There is no common ratio meaning it is not a geometric sequence. Although the sequence is neither arithmetic nor geometric, we can write a recursive rule for it. Let's look for a pattern in the sum of consecutive terms. a1+a2a2+a3a3+a4=2+4 =6=4+6 =10=6+10=16 We see that beginning with the third term each term is the sum of two previous terms. Hence, a recursive equation for the sequence is an=an−2+an−1. The first two terms and the recursive equation gives us the recursive rule. Recursive Rule:a1=2, a2=4, an=an−2+an−1 | |

Exercises 47 We are asked to write the first 2nd, 5th and 10th term of a sequence, given a recursive rule. f(1)f(n)=3=f(n−1)+7, for n>1 To do so, we will use a table.nf(n)=f(n−1)+7f(n−1)+7f(n) 1f(1)=3−3 2f(2)=f(2−1)+7f(1)+7=3+710 3f(3)=f(3−1)+7f(2)+7=10+717 4f(4)=f(4−1)+7f(3)+7=17+724 5f(5)=f(5−1)+7f(4)+7=24+731 6f(6)=f(6−1)+7f(5)+7=31+738 7f(7)=f(7−1)+7f(6)+7=38+745 8f(8)=f(8−1)+7f(7)+7=45+752 9f(9)=f(9−1)+7f(8)+7=52+759 10f(10)=f(10−1)+7f(9)+7=59+766 Therefore, f(2)=10, f(5)=31 and f(10)=66. | |

Exercises 48 We are asked to write the first 2nd, 5th and 10th term of a sequence, given a recursive rule. f(1)f(n)=-1=6f(n−1), for n>1 To do so, we will use a table.nf(n)=6f(n−1)6f(n−1)f(n) 1f(1)=-1−-1 2f(2)=6f(2−1)6f(1)=6(-1)-6 3f(3)=6f(3−1)6f(2)=6(-6)-36 4f(4)=6f(4−1)6f(3)=6(-36)-216 5f(5)=6f(5−1)6f(4)=6(-216)-1296 6f(6)=6f(6−1)6f(5)=6(-1296)-7776 7f(7)=6f(7−1)6f(6)=6(-7776)-46656 8f(8)=6f(8−1)6f(7)=6(-46656)-279936 9f(9)=6f(9−1)6f(8)=6(-279936)-1679616 10f(10)=6f(10−1)6f(9)=6(-1679616)-10077696 Therefore, f(2)=-6, f(5)=-1296 and f(10)=-10077696. | |

Exercises 49 We are asked to write the first 2nd, 5th and 10th term of a sequence, given a recursive rule. f(1)f(n)=8=-f(n−1), for n>1 To do so, we will use a table.nf(n)=-f(n−1)-f(n−1)f(n) 1f(1)=8−8 2f(2)=-f(2−1)-f(1)=-8-8 3f(3)=-f(3−1)-f(2)=-(-8)8 4f(4)=-f(4−1)-f(3)=-8-8 5f(5)=-f(5−1)-f(4)=-(-8)8 6f(6)=-f(6−1)-f(5)=-8-8 7f(7)=-f(7−1)-f(6)=-(-8)8 8f(8)=-f(8−1)-f(7)=-8-8 9f(9)=-f(9−1)-f(8)=-(-8)8 10f(10)=-f(10−1)-f(9)=-8-8 Therefore, f(2)=-8, f(5)=8 and f(10)=-8. | |

Exercises 50 We are asked to write the first 2nd, 5th and 10th term of a sequence, given a recursive rule. f(1)f(2)f(n)=4=5=f(n−1)+f(n−2), for n>2 To do so, we will use a table.nf(n)=f(n−1)+f(n−2)f(n−1)+f(n−2)f(n) 1f(1)=4−4 2f(2)=5−5 3f(3)=f(3−1)+f(3−2)f(2)+f(1)=5+49 4f(4)=f(4−1)+f(4−2)f(3)+f(2)=9+514 5f(5)=f(5−1)+f(5−2)f(4)+f(3)=14+923 6f(6)=f(6−1)+f(6−2)f(5)+f(4)=23+1437 7f(7)=f(7−1)+f(7−2)f(6)+f(5)=37+2360 8f(8)=f(8−1)+f(8−2)f(7)+f(6)=60+3797 9f(9)=f(9−1)+f(9−2)f(8)+f(7)=97+60157 10f(10)=f(10−1)+f(10−2)f(9)+f(8)=157+97254 Therefore, f(2)=5, f(5)=23 and f(10)=254. | |

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