Download 10-1 Sequences, Series, and Sigma Notation

10-1 Sequences, Series, and Sigma Notation
Find the next four terms of each sequence.
1. 1, 8, 15, 22, …
SOLUTION: The terms appear to increase by 7. Check.
The next four terms are:
.
2. 3, −6, 12, −24, …
SOLUTION: These terms appear to be multiplied by –2. Check.
The next four terms are:
.
3. 81, 27, 9, 3, …
SOLUTION: These terms appear to be divided by 3. Check.
The next four terms are:
.
4. 1, 3,
7, 13,- Powered
…
eSolutions
Manual
by Cognero
SOLUTION: If we subtract each term from the term that follows, we see a pattern.
Page 1
10-1 Sequences, Series, and Sigma Notation
.
4. 1, 3, 7, 13, …
SOLUTION: If we subtract each term from the term that follows, we see a pattern.
3 – 1 = 2
7– 3=4
13 – 7 = 6
It appears that each term is generated by adding the next successive even number. So the next four terms are:
13 + 8 = 21
21 + 10 = 31
31 + 12 = 43
43 + 14 = 57.
5. −2, −15, −28, −41…
SOLUTION: If we subtract each term from the term that follows, we see a pattern.
–15 – (–2) = –13
–28 – (–15) = –13
–41 – (–28) = –13
Each term is 13 less than the previous term. The next four terms are –41 + (–13) = –54
–54– (–13) = –67
–67 – (–13) = –80
–80– (–13) = –93.
6. 1, 4, 10, 19, …
SOLUTION: If we subtract each term from the term that follows, we see a pattern.
4– 1=3
10 – 4 = 6
19 – 10 = 9
It appears that each term is generated by adding the next successive multiple of 3. The next four terms are :
19 + 12 = 31
31+ 18 = 46
46 + 21 = 64
64 + 24 = 85.
Find the first four terms of each sequence.
7. a n = n 2 – 1
SOLUTION: 2
a 1 = 1 – 1 = 0
2
a 2 = 2 – 1 = 3
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Manual - Powered by Cognero
2
a3 = 3 – 1
= 8
Page 2
19 + 12 = 31
31+ 18 = 46
+ 21 = 64
10-1 46
Sequences,
64 + 24 = 85.
Series, and Sigma Notation
Find the first four terms of each sequence.
7. a n = n 2 – 1
SOLUTION: 2
a 1 = 1 – 1 = 0
2
a 2 = 2 – 1 = 3
2
a3 = 3 – 1
= 8
2
a 4 = 4 – 1
= 15
8. a n = −2n + 7
SOLUTION: 1
a 1 = −2 + 7
= –2 + 7 = 5
2
a 2 = −2 + 7
= –4 + 7 = 3
3
a 3 = −2 + 7
= –8 + 7 = –1
4
a 4 = −2 + 7
= –16 + 7 = –9
9. a n =
SOLUTION: eSolutions Manual - Powered by Cognero
Page 3
= –1
4
a 4 = −2 + 7
10-1 = –16 + 7 Sequences,
= –9
Series, and Sigma Notation
9. a n =
SOLUTION: 10. a n = (–1)n + 1 + n
SOLUTION: 1+1
a 1 = (–1)
+1
2
= 1 + 1
= 2
2+1
a 2 = (–1)
+2
3
= (–1) + 2 = 1
3+1
a 3 = (–1)
+3
4
= (–1) + 3 = 4
4+1
a 4 = (–1)
+4
5
= (–1) + 4 = 3
11. AUTOMOBILE LEASES Lease agreements often contain clauses that limit the number of miles driven per year
by charging a per-mile fee over that limit. For the car shown below, the lease requires that the number of miles
driven each year must be no more than 15,000.
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Page 4
= 4
4+1
a 4 = (–1)
+4
5
+ 4 Series, and Sigma Notation
10-1 = (–1)
Sequences,
= 3
11. AUTOMOBILE LEASES Lease agreements often contain clauses that limit the number of miles driven per year
by charging a per-mile fee over that limit. For the car shown below, the lease requires that the number of miles
driven each year must be no more than 15,000.
a. Write the sequence describing the maximum number of allowed miles on the car at the end of every 12 months
of the lease if the car has 1350 miles at the beginning of the lease.
b. Write the first 4 terms of the sequence that gives the cumulative cost of the lease for a given month.
c. Write an explicit formula to represent the sequence in part b.
d. Determine the total amount of money paid by the end of the lease.
SOLUTION: a. If the car starts with 1350 miles on it and the number of additional miles must be less than 15,000, for the
sequence by adding 15,000 to the previous term.
a 1 = 1350 + 15,000 or 16,350
a 2 = 16,350 + 15,000 or 31,350
a 3 = 31,350 + 15,000 or 46,350
b. During the first month, the customer pays the down payment plus the first month’s lease payment. After that,
the customer pays just the lease payment each month.
b 1 = 1699 + 399 or 2098
b 2 = 2098 + 399 or 2497
b 3 = 2497 + 399 or 2896
b 4 = 2896 + 399 or 3295
So the first four terms of the sequence are 2098, 2497, 2896, and 3295.
c. The sequence in part b is 2098, 2497, 2896, 3295. Each term a n in this sequence can be found by adding multiple
n of 399 to an initial value of 1699. So an explicit formula for this sequence is a n = 1699 + 399n.
d. The lease is for 36 months. Use the explicit formula you wrote in part a to find a 36.
a 36 = 1699 + 399(36) = 16,063
The total amount of money paid by the end of the lease is $16,063.
Find the specified term of each sequence.
12. 4th term, a 1 = 5, a n = –3a n – 1 + 10, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 4th term a 4 must be found
first.
a =5
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Page 5
a 2 = –3a 2 – 1 + 10
d. The lease is for 36 months. Use the explicit formula you wrote in part a to find a 36.
= 16,063
36 = 1699 + 399(36)
10-1 aSequences,
Series,
and Sigma Notation
The total amount of money paid by the end of the lease is $16,063.
Find the specified term of each sequence.
12. 4th term, a 1 = 5, a n = –3a n – 1 + 10, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 4th term a 4 must be found
first.
a1 = 5
a 2 = –3a 2 – 1 + 10
= –3a 1 + 10
= –3(5) + 10
= –5
a 3 = –3a 3 – 1 + 10
= –3a 2 + 10
= –3(–5) + 10
= 25
a 4 = –3a 4 – 1 + 10
= –3a 3 + 10
= –3(25) + 10
= –65
13. 7th term, a 1 = 14, a n = 0.5a n − 1 + 3, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 7th term a 7 must be found
first.
a 1 = 14
a 2 = 0.5a 2 − 1 + 3
= 0.5a 1 + 3
= 0.5(14) + 3 = 10
a 3 = 0.5a 3 – 1 + 3
= 0.5a 2 + 3
= 0.5(10) + 3 = 8
a 4 = 0.5a 4 – 1 + 3
= 0.5a 3 + 3
= 0.5(8) + 3 = 7
a 5 = 0.5a 5 – 1 + 3
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= 0.5a 4 + 3
= 0.5(7) + 3 = 6.5
Page 6
a 4 = –3a 4 – 1 + 10
= –3a 3 + 10
= –3(25) + 10
10-1 = –65
Sequences, Series, and Sigma Notation
13. 7th term, a 1 = 14, a n = 0.5a n − 1 + 3, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 7th term a 7 must be found
first.
a 1 = 14
a 2 = 0.5a 2 − 1 + 3
= 0.5a 1 + 3
= 0.5(14) + 3 = 10
a 3 = 0.5a 3 – 1 + 3
= 0.5a 2 + 3
= 0.5(10) + 3 = 8
a 4 = 0.5a 4 – 1 + 3
= 0.5a 3 + 3
= 0.5(8) + 3 = 7
a 5 = 0.5a 5 – 1 + 3
= 0.5a 4 + 3
= 0.5(7) + 3 = 6.5
a 6 = 0.5a 6 – 1 + 3
= 0.5a 5 + 3
= 0.5(6.5) + 3 = 6.25
a 7 = 0.5a 7 – 1 + 3
= 0.5a 7 + 3
= 0.5(6.25) + 3 = 6.125
14. 4th term, a 1 = 0,
, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 4th term a 4 must be found
first.
a1 = 0
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Page 7
a 7 = 0.5a 7 – 1 + 3
= 0.5a 7 + 3
10-1 = 0.5(6.25) + 3 Sequences, Series,
= 6.125
14. 4th term, a 1 = 0,
and Sigma Notation
, n ≥ 2
SOLUTION: This formula given is a recursive one. This requires that each of the terms before the 4th term a 4 must be found
first.
a1 = 0
15. 3rd term, a 1 = 3, a n = (a n – 1)2 – 5a n – 1 + 4, n ≥ 2
SOLUTION: 2
3rd term, a n = (a n – 1) – 5a n – 1 + 4, a 1 = 3
This formula given is a recursive one. This requires that each of the terms before the 3th term a 3 must be found
first.
a1 = 3
2
a 2 = (a 2 – 1) – 5a 2 – 1 + 4
2
= (a 1) – 5a 1 + 4
2
= 3 – 5(3) + 4
= 9 – 15 + 4 = –2
2
a 3 = (a 3 – 1) – 5a 3 – 1 + 4
2
= (a 2) – 5a 2 + 4
2
= (–2) – 5(–2) + 4
= 4 + 10 + 4 = 18
16. WEB SITE Khari, the student from the beginning of the lesson, had great success expanding her Web site. Each
student who received a referral developed a Web page and referred five more students to Khari’s site.
a sequence modeling the number of new Web pages created through Khari’s site.Page 8
b. Suppose the school has 1576 students. After how many rounds of referrals did the entire student body have a
Web page?
eSolutions
by terms
Cognero
a. Manual
List the- Powered
first five
of
2
= (a 2) – 5a 2 + 4
2
= (–2) – 5(–2) + 4
10-1 = 4 + 10 + 4 Sequences, Series,
= 18
and Sigma Notation
16. WEB SITE Khari, the student from the beginning of the lesson, had great success expanding her Web site. Each
student who received a referral developed a Web page and referred five more students to Khari’s site.
a. List the first five terms of a sequence modeling the number of new Web pages created through Khari’s site.
b. Suppose the school has 1576 students. After how many rounds of referrals did the entire student body have a
Web page?
SOLUTION: a. An explicit formula for this sequence is a n = 5n . Use this formula to find the first five terms of the sequence, a 0,
a 1, a 2, a 3, and a 4, where n represents the first round of referrals.
0
a0 = 5 = 1
1
a1 = 5 = 5
2
a 2 = 5 = 25
3
a 3 = 5 = 125
4
a 4 = 5 = 625
b. To determine after how many rounds of referrals the entire student body had a Web page, find one or two more
terms of the sequence. Then find the cumulative sum of the sequence after each term.
5
a 5 = 5 = 3125
6
a 6 = 5 = 15,625
a0 + a1 = 1 + 5 = 6
a 0 + a 1 + a 2 = 6 + 25 = 31
a 0 + a 1 + a 2 + a 3 = 31 + 125 = 156
a 0 + a 1 + a 2 + a 3 + a 4 = 156 + 625 = 781
a 0 + a 1 + a 2 + a 3 + a 4 + a 5 = 781 + 3125 = 3206
a 0 + a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 3026 + 15,625 = 18,651
After 5 rounds of referrals, then total number of new Web pages is greater than the population of the school since
2306 > 1576.
17. BEES Female honeybees come from fertilized eggs (male and female parent), while male honeybees come from
unfertilized eggs (only one female parent).
a. Draw a family tree showing the 3 previous generations of a male honeybee (parents only).
b. Determine the number of parent bees in the 11th previous generation of a male honeybee.
SOLUTION: a. Start with a male honeybee in generation 4 and work backward. A male honeybee can only have a female
parent.
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A female honeybee has a male and a female parent.
Page 9
0
1
2
3
4
5
a 0 + a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 3026 + 15,625 = 18,651
5 rounds ofSeries,
referrals,and
then total
number
of new Web pages is greater than the population of the school since
10-1 After
Sequences,
Sigma
Notation
2306 > 1576.
17. BEES Female honeybees come from fertilized eggs (male and female parent), while male honeybees come from
unfertilized eggs (only one female parent).
a. Draw a family tree showing the 3 previous generations of a male honeybee (parents only).
b. Determine the number of parent bees in the 11th previous generation of a male honeybee.
SOLUTION: a. Start with a male honeybee in generation 4 and work backward. A male honeybee can only have a female
parent.
A female honeybee has a male and a female parent.
Again, a male honeybee can only have a female parent, and a female honeybee has a male and a female parent.
b. Let Bn be the number of bees in the nth previous level of the tree, with level 0 corresponding to the original male
bee. Each of the B n − 1 bees at level n − 1 will have a mother, so there are Bn − 1 female bees at level n. Each
female at level n − 1 will also have a father and each of these is a mother to exactly one of the bees at level n − 2,
so there are Bn − 2 male bees at level n. Thus, Bn = Bn − 1 + Bn − 2 with B0 = 1 and B1 = 1. This is the Fibonacci
sequence, in which the previous two terms are added to get the next term. Find the 11th term in the Fibonacci
sequence.
4th term: 2 + 3 = 5
5th term: 3 + 5 = 8
6th term: 5 + 8 = 13
7th term: 8 + 13 = 21
8th term: 13 + 21 = 34
9th term: 21 + 34 = 55
10th term: 34 + 55 = 89
11th term: 55 + 89 = 144
Therefore, there will be 144 parent bees in the 11th previous generation.
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by Cognero
Determine
whether
each
18. a 1 = 4, 1.5a n – 1, n
2
sequence is convergent or divergent.
Page 10
8th term: 13 + 21 = 34
9th term: 21 + 34 = 55
10th term: 34 + 55 = 89
term: 55 + 89
= 144 and Sigma Notation
10-1 11th
Sequences,
Series,
Therefore, there will be 144 parent bees in the 11th previous generation.
Determine whether each sequence is convergent or divergent.
18. a 1 = 4, 1.5a n – 1, n 2
SOLUTION: The first term in this sequence is 4. Find several more terms using the given recursive formula.
The first eight terms of the sequence are 4, 6, 9, 13.5, 20.25, 30.375, 45.5625, and 68.34375. These terms do not
appear to approach a finite number. Therefore, the sequence is divergent.
19. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
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Page 11
The first eight terms of the sequence are 4, 6, 9, 13.5, 20.25, 30.375, 45.5625, and 68.34375. These terms do not
10-1 appear
Sequences,
Series,
Sigma
Notation
to approach
a finiteand
number.
Therefore,
the sequence is divergent.
19. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
The first eight terms of this sequence are 0.5, 0.05, 0.005, 0.0005, 0.00005, 0.000005, 0.0000005, and 0.00000005.
These terms appear to slowly approach a finite number, 0. The sequence appears to have a limit and is therefore
convergent.
20. a n = –n 2 – 8n + 106
SOLUTION: Find several terms in the sequence using the given explicit formula.
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Page 12
The first eight terms of this sequence are 0.5, 0.05, 0.005, 0.0005, 0.00005, 0.000005, 0.0000005, and 0.00000005.
terms appear
to slowly
approach
number, 0. The sequence appears to have a limit and is therefore
10-1 These
Sequences,
Series,
and
Sigmaa finite
Notation
convergent.
20. a n = –n 2 – 8n + 106
SOLUTION: Find several terms in the sequence using the given explicit formula.
The first eight terms of this sequence are 97, 86, 73, 58, 41, 41, 22, 1, and –22. These terms do not appear to
approach a finite number. Therefore, the sequence is divergent.
21. a 1 = –64,
a n – 1, n
2
SOLUTION: The first term in this sequence is –64. Find several more terms using the given recursive formula.
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Page 13
first eight terms
of thisand
sequence
are 97,
86, 73, 58, 41, 41, 22, 1, and –22. These terms do not appear to
10-1 The
Sequences,
Series,
Sigma
Notation
approach a finite number. Therefore, the sequence is divergent.
21. a 1 = –64,
2
a n – 1, n
SOLUTION: The first term in this sequence is –64. Find several more terms using the given recursive formula.
The first eight terms of this sequence are –64, –48, –36, –27, –20.25, –15.1875, –11.390625, and –8.54296875.
These terms appear to slowly approach a finite number, 0. The sequence appears to have a limit and is therefore
convergent.
22. a 1 = 1,
n
2
SOLUTION: The first term in this sequence is 1. Find several more terms using the given recursive formula.
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Page 14
The first eight terms of this sequence are –64, –48, –36, –27, –20.25, –15.1875, –11.390625, and –8.54296875.
terms appear
to slowly
approach
number, 0. The sequence appears to have a limit and is therefore
10-1 These
Sequences,
Series,
and
Sigmaa finite
Notation
convergent.
22. a 1 = 1,
n
2
SOLUTION: The first term in this sequence is 1. Find several more terms using the given recursive formula.
The first eight terms of this sequence are 1, 3, 1, 3, 1, 3, 1, and 3. It appears that when n is odd, a n is 1 and when n
is even, a n is 3. Since a n does not approach one particular value, the sequence has no limit. Therefore, the
sequence is divergent.
23. a n = n 2 – 3n + 1
SOLUTION: Find several terms in the sequence using the given explicit formula.
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Page 15
The first eight terms of this sequence are 1, 3, 1, 3, 1, 3, 1, and 3. It appears that when n is odd, a n is 1 and when n
even, a n is 3. Since a n does not approach one particular value, the sequence has no limit. Therefore, the
10-1 isSequences,
Series, and Sigma Notation
sequence is divergent.
23. a n = n 2 – 3n + 1
SOLUTION: Find several terms in the sequence using the given explicit formula.
These terms are increasing and do not approach a finite number. Therefore, the sequence is divergent.
24. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
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These terms are increasing and do not approach a finite number. Therefore, the sequence is divergent.
Page 16
10-1 Sequences, Series, and Sigma Notation
These terms are increasing and do not approach a finite number. Therefore, the sequence is divergent.
24. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
These terms are increasing and do not approach a finite number. Therefore, the sequence is divergent.
25. a 1 = 9, a n =
,n
2
SOLUTION: The first term in this sequence is 9. Find several more terms using the given recursive formula.
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Page 17
10-1 Sequences, Series, and Sigma Notation
These terms are increasing and do not approach a finite number. Therefore, the sequence is divergent.
25. a 1 = 9, a n =
,n
2
SOLUTION: The first term in this sequence is 9. Find several more terms using the given recursive formula.
These terms slowly approach a finite number, 3. The sequence appears to have a limit and is therefore convergent.
26. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
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Page 18
10-1 Sequences, Series, and Sigma Notation
These terms slowly approach a finite number, 3. The sequence appears to have a limit and is therefore convergent.
26. a n =
SOLUTION: Find several terms in the sequence using the given explicit formula.
These terms slowly approach a finite number, 5. The sequence appears to have a limit and is therefore convergent.
27. SOLUTION: Find several terms in the sequence using the given explicit formula.
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Page 19
27. 10-1 SOLUTION: Sequences, Series, and Sigma Notation
Find several terms in the sequence using the given explicit formula.
The first eight terms of this sequence are 2, 1.4, 1.12, 1.032, 1.008, 1.00195, 1.000448, and 1.0001024. These terms
approach a finite number, 1. The sequence appears to have a limit and is therefore convergent.
Find the indicated sum for each sequence.
4)(n – 3)
n
eSolutions
28. 5thManual
partial- Powered
sum of by
a Cognero
= n(n –
SOLUTION: Page 20
first eight terms
of thisand
sequence
are 2,Notation
1.4, 1.12, 1.032, 1.008, 1.00195, 1.000448, and 1.0001024. These terms
10-1 The
Sequences,
Series,
Sigma
approach a finite number, 1. The sequence appears to have a limit and is therefore convergent.
Find the indicated sum for each sequence.
28. 5th partial sum of a n = n(n – 4)(n – 3)
SOLUTION: Find the first five terms of the sequence.
The 5th partial sum of this sequence is 6 + 4 + 0 + 0 + 10 is 20
29. 6th partial sum of a n =
SOLUTION: Find the first six terms of the sequence.
The 6th partial sum of this sequence is –2 + (–3.5) + (–4) + (–4.25) + (–4.4) + (–4.5) or –22.65.
30. S 8 of a 1 = 1, a n = a n – 1 + (18 – n), n
2
SOLUTION: Analyze the pattern by finding the differences between successive terms.
18 – 1 = 17
34 – 18 = 16
49 – 34 = 15
eSolutions
by Cognero
63Manual
– 49 =- Powered
14
The sequence is formed by adding successively smaller integers.
Continue this pattern to find the next three terms.
Page 21
10-1 Sequences, Series, and Sigma Notation
The 6th partial sum of this sequence is –2 + (–3.5) + (–4) + (–4.25) + (–4.4) + (–4.5) or –22.65.
30. S 8 of a 1 = 1, a n = a n – 1 + (18 – n), n
2
SOLUTION: Analyze the pattern by finding the differences between successive terms.
18 – 1 = 17
34 – 18 = 16
49 – 34 = 15
63 – 49 = 14
The sequence is formed by adding successively smaller integers.
Continue this pattern to find the next three terms.
63 + 13 = 76
76 + 12 = 88
88 + 11 = 99
The 8th partial sum S 8 of this sequence is 1 + 18 + 34 + 49 + 63 + 76 + 88 + 99 or 428.
31. S 4 of a 1 = 64, a n =
a n – 1, n
2
SOLUTION: Find the next three terms of the sequence.
The 4th partial sum S 4 of this sequence is 64 + (–48) + 36 + (–27) or 25.
32. 11th partial sum of a 1 = 4,
SOLUTION: The first term of the sequence is 4. Find the next 10 terms of the sequence.
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Page 22
1
SOLUTION: The first term of the sequence is 4. Find the next 10 terms of the sequence.
10-1 Sequences, Series, and Sigma Notation
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Page 23
10-1 Sequences, Series, and Sigma Notation
The 11th partial sum of this sequence is 4 + (–7) + 10 + (–13) + 16 + (–19) + 22 + (–25) + 28 + (–31) + 34 or 19.
33. S 9 of a 1 = –35, a n = a n – 1 + 8, n
2
SOLUTION: The first term of the sequence is –35. Find the next 8 terms of the sequence.
The 9th partial sum S of this sequence is –35 + (–27) + (–19) + (–11) + (–3) + 5 + 13 + 21 + 29 or –27.
9
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34. 4th partial sum of a 1 = 3, a n = (a n – 1 – 2)3, n ≥ 2
Page 24
10-1 Sequences, Series, and Sigma Notation
The 11th partial sum of this sequence is 4 + (–7) + 10 + (–13) + 16 + (–19) + 22 + (–25) + 28 + (–31) + 34 or 19.
33. S 9 of a 1 = –35, a n = a n – 1 + 8, n
2
SOLUTION: The first term of the sequence is –35. Find the next 8 terms of the sequence.
The 9th partial sum S 9 of this sequence is –35 + (–27) + (–19) + (–11) + (–3) + 5 + 13 + 21 + 29 or –27.
34. 4th partial sum of a 1 = 3, a n = (a n – 1 – 2)3, n ≥ 2
SOLUTION: eSolutions Manual - Powered by Cognero
Find the next three terms of the sequence.
Page 25
10-1 The
Sequences,
Series,
and
Sigma
Notation
9th partial sum
S of this
sequence
is –35
+ (–27) + (–19) + (–11) + (–3) + 5 + 13 + 21 + 29 or –27.
9
34. 4th partial sum of a 1 = 3, a n = (a n – 1 – 2)3, n ≥ 2
SOLUTION: Find the next three terms of the sequence.
The 4th partial sum of this sequence is 3 + 1 + (–1) + (–27) or –24.
35. S 4 of a n =
SOLUTION: Find the first 4 terms of the sequence.
The 4th partial sum of this sequence is
which is Find each sum.
36. eSolutions Manual - Powered by Cognero
SOLUTION: Page 26
10-1 The
Sequences,
Series,
Sigma
Notation
4th partial sum
of this and
sequence
is
which is Find each sum.
36. SOLUTION: 37. SOLUTION: 38. SOLUTION: 39. SOLUTION: 40. eSolutions Manual - Powered by Cognero
SOLUTION: Page 27
10-1 Sequences, Series, and Sigma Notation
40. SOLUTION: 41. SOLUTION: 42. SOLUTION: 43. SOLUTION: eSolutions Manual - Powered by Cognero
Page 28
10-1 Sequences, Series, and Sigma Notation
43. SOLUTION: 44. SOLUTION: 45. SOLUTION: 46. FINANCIAL LITERACY Jim’s bank account had an initial deposit of $380, earning 3.5% interest per year
compounded annually.
a. Find the balance each year for the first five years.
b. Write a recursive and an explicit formula defining his account balance.
c. For very large values of n, which formula gives a more accurate balance? Explain.
SOLUTION: a. The compound interest formula is
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. In this situation, P = 380, r = 0.035, and n = 1, so the formula
Page 29
becomes
, where t is the number of years after the initial deposit. Use this equation to find the first
five terms of the sequence.
10-1 Sequences, Series, and Sigma Notation
46. FINANCIAL LITERACY Jim’s bank account had an initial deposit of $380, earning 3.5% interest per year
compounded annually.
a. Find the balance each year for the first five years.
b. Write a recursive and an explicit formula defining his account balance.
c. For very large values of n, which formula gives a more accurate balance? Explain.
SOLUTION: a. The compound interest formula is
. In this situation, P = 380, r = 0.035, and n = 1, so the formula
becomes
, where t is the number of years after the initial deposit. Use this equation to find the first
five terms of the sequence.
b. In this sequence, each term is 1.035 times the previous term. Therefore, a recursive formula for this sequence is
a 0 = 380 and a t = 1.035a t – 1, t 1, where t is the number of years after the initial deposit.
t
An explicit formula for this sequences is a t = 380(1.035) , where t is the number of years after the initial deposit.
This is the same formula as that given by the compound interest formula for this situation, replacing P with a t.
c. Explicit formula; when the recursive formula is used, rounding occurs at each step. With an explicit formula, you
only need to round the final answer.
47. INVESTING Melissa invests $200 every 3 months. The investment pays an annual percentage rate of 8% and the
interest is compounded quarterly. If Melissa makes each payment at the beginning of the quarter and the interest is
posted at the end of the quarter, what will the total value of the investment be after 2 years?
SOLUTION: Since the interest is compounded quarterly, divide the rate 0.08 by 4 to get a quarterly rate of 0.02. So the value of
the investment at the end of the first quarter is 200 + 200(0.02), which is 200(1 + 0.02) or 200(1.02) or 204.
Melissa continues to add money to the account each quarter. Therefore, we can define the sequence of quarterly
,
investment values as
, where n 2.
Two years is equivalent to 2 × 4 or 8 quarters. Therefore, you need to find the 8th term in this recursively defined
sequence.
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Page 30
10-1 Sequences, Series, and Sigma Notation
After 2 years or 8 quarters, the total value of the investment will be about $1750.93.
48. RIDES The table shows the number of riders of the Mean Streak each year from 1998 to 2007. This ridership
data can be approximated by
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, where n = 1 represents 1998, n = 2 represents 1999, and so on.
Page 31
After 2 years or 8 quarters, the total value of the investment will be about $1750.93.
48. RIDES The table shows the number of riders of the Mean Streak each year from 1998 to 2007. This ridership
10-1 Sequences, Series, and Sigma Notation
data can be approximated by
, where n = 1 represents 1998, n = 2 represents 1999, and so on.
a. Sketch a graph of the number of riders from 1998 to 2007. Then determine whether the sequence appears to be
convergent or divergent. Does this make sense in the context of the situation? Explain your reasoning.
b. Use the table to find the total number of riders from 1998 to 2005. Then use the explicit sequence to find the 8th
partial sum of a n. Compare the results.
c. If the sequence continues, find a14. What does this number represent?
SOLUTION: a.
Divergent; the sequence appears to continue to decrease. It could not decrease forever because the number of
riders cannot be less than 0.
b. Using the table, the total number of riders from 1998 to 2005 is 1.31 + 1.15 + 1.14 + 1.09 + 1.05 + 0.99 + 0.95 +
0.89 or 8.57 million. Now fine the 8th partial sum of the sequence given by the explicit formula
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.
Page 32
10-1 Sequences, Series, and Sigma Notation
The 8th partial sum of this sequence is 1.25 + 1.2 + 1.15 + 1.1 + 1.05 + 1 + 0.95 + 0.9 or 8.6 million. The results
are approximately equal.
c.
a 14 represents the number of riders in 2011, 0.6 million riders.
Copy and complete the table.
49. Recursive Formula
Explicit
Formula
Sequence
6, 8, 10, 12, …
SOLUTION: Sample answer: The first term is 6 and each successive term is 2 more than the term before it, so a 1 = 6, a n = a n –
1
+ 2, for n
2.
Since 2(1) + 4 = 6, 2(2) + 4 = 8, 2(3) + 4 = 10, and 2(4) + 4 = 12, then a n = 2n + 4
50. Recursive Formula
a 1 = 15, a n = a n – 1 – 1, n
Explicit
Formula
Sequence
2
SOLUTION: a 1 = 15, a 2 = 15 – 1 = 14, a 3 = 14 – 1 = 13, a 4 = 13 – 1 = 12, so the sequence is 15, 14, 13, 12, …;
Since 15 = –1 + 16, 14 = –2 + 16, 13 = –3 + 16, and 12 = –4 + 16, then a n = –n + 16.
51. Recursive Formula
Sequence
Explicit
Formula
7, 21, 63, 189, …
SOLUTION: Sample answer: The first term is 7 and each successive term is 3 times larger than the term before it, so a 1 = 7, a n
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= 3a n – 1, for n
Page 33
2.
2
3
n –1
Since 7 = 7(1), 21 = 7(3), 63 = 7(3 ), and 189 = 7(3 ), then a n = 7(3)
SOLUTION: a 1 = 15, a 2 = 15 – 1 = 14, a 3 = 14 – 1 = 13, a 4 = 13 – 1 = 12, so the sequence is 15, 14, 13, 12, …;
10-1 Since
Sequences,
Series,
15 = –1 + 16,
14 = –2and
+ 16,Sigma
13 = –3 Notation
+ 16, and 12 = –4 + 16, then a
n
= –n + 16.
51. Recursive Formula
Sequence
Explicit
Formula
7, 21, 63, 189, …
SOLUTION: Sample answer: The first term is 7 and each successive term is 3 times larger than the term before it, so a 1 = 7, a n
= 3a n – 1, for n
2.
2
n –1
3
Since 7 = 7(1), 21 = 7(3), 63 = 7(3 ), and 189 = 7(3 ), then a n = 7(3)
52. Recursive
Formula
Sequence
Explicit Formula
a n = 10(–2)n
SOLUTION: 1
2
3
4
Since a 1 = 10(–2) = –20, a 2 = 10(–2) = 40, a 3 = 10(–2) = –80, and a 4 = 10(–2) = 160, the sequence is –20, 40,
–80, 160, …;
The first term is –20 and each successive term is –2 times larger than the term before it, so a 1 = –20 and a n = –
2a n – 1, for n
2.
53. Recursive Formula
Explicit
Formula
Sequence
a n = 8n – 3
SOLUTION: Since a 1 = 8(1) – 3 = 5, a 2 = 8(2) – 3 = 13, a 3 = 8(3) – 3 = 21, and a 4 = 8(4) – 3 = 29, the sequence is 5, 13, 21,
29, …;
The first term is 5 and each successive term is 8 more than the term before it, so a 1 = 5 and a n = a n – 1 + 8, for n
2;
54. Recursive Formula
a 1 =2, a n = 4a n – 1 , n
Explicit
Formula
Sequence
2
SOLUTION: Since a 1 = 2, a 2 = 4(2) = 8, a 3 = 4(8) = 32, and a 4 = 4(32) = 128, the sequence is 2, 8, 32, 128, …;
2
3
n–1
Since 2 = 2(1), 8 = 2(4), 32 = 2(4 ), and 128 = 2(4 ), then a n = 2(4
).
55. Recursive Formula
eSolutionsaManual
=3, a n- =Powered
a n – 1 +by
2nCognero
– 1, n
1
SOLUTION: Explicit
Formula
Sequence
2
Page 34
SOLUTION: Since a 1 = 2, a 2 = 4(2) = 8, a 3 = 4(8) = 32, and a 4 = 4(32) = 128, the sequence is 2, 8, 32, 128, …;
2
3
10-1 Since
Sequences,
Notation
2 = 2(1), 8Series,
= 2(4), 32and
= 2(4Sigma
), and 128
= 2(4 ), then a
n–1
).
n = 2(4
55. Recursive Formula
a 1 =3, a n = a n – 1 + 2n – 1, n
Explicit
Formula
Sequence
2
SOLUTION: Since a 1 = 3, a 2 = 3 + 2(2) – 1 = 6, a 3 = 6 + 2(3) – 1 = 11, and a 4 = 11 + 2(4) – 1 = 18, the sequence is 3, 6, 11,
18, …;
2
2
2
2
2
Since 3 = 1 + 2, 6 = 2 + 2, 11 = 3 + 2, and 18 = 4 + 2, then a n = n + 2.
56. Recursive Formula
Sequence
Explicit
Formula
an = n2 + 1
SOLUTION: 2
2
2
2
Since a 1 = 1 + 1 = 2, a 2 = 2 + 1 = 5, a 3 = 3 + 1 = 10, and a 4 = 4 + 1 = 17, the sequence is 2, 5, 10, 17, …;
The first term is 2 and each successive term is 2n – 1 more than the term before it, so a 1 = 2 and a n = a n – 1 + 2n
– 1, for n
2.
Write each series in sigma notation. The lower bound is given.
57. –2 – 1 + 0 + 1 + 2 + 3 + 4 + 5; n = 1
SOLUTION: 58. + +
+ + + ;n=4
SOLUTION: 59. 8 + 27 + 64 + … + 1000; n = 2
SOLUTION: eSolutions Manual - Powered by Cognero
60. +
+ + … +
;n=1
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10-1 Sequences, Series, and Sigma Notation
59. 8 + 27 + 64 + … + 1000; n = 2
SOLUTION: 60. +
+ + … +
;n=1
SOLUTION: 61. –8 + 16 – 32 + 64 – 128 + 256 – 512; n = 3
SOLUTION: 62. ;n=1
SOLUTION: Determine whether each sequence is convergent or divergent. Then find the fifth partial sum of the
sequence.
63. SOLUTION: Find the first five terms of the sequence using the explicit formula,
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.
Page 36
10-1 Sequences, Series, and Sigma Notation
Determine whether each sequence is convergent or divergent. Then find the fifth partial sum of the
sequence.
63. SOLUTION: Find the first five terms of the sequence using the explicit formula,
.
These terms do not approach a finite number. Therefore, the sequence is divergent.
The sum of the first five terms is 1 + 0 + (–1) + 0 + 1 or 1.
64. SOLUTION: Find the first five terms of the sequence using the explicit formula,
.
These terms continue to decrease; they do not approach a finite number. Therefore, the sequence is divergent.
The sum of the first five terms is (–1) + (–2) + (–3) + (–4) + (–5) or –15.
65. SOLUTION: Find the first five terms of the sequence using the explicit formula,
.
These terms approach a finite number, 0. Therefore, the sequence is convergent.
The approximate sum of the first five terms is (–0.61) + (0.37) + (–0.22) + (0.14) + (–0.08) or about –0.4.
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Page 37
66. WATER PRESSURE The pressure exerted on the human body at sea level is 14.7 pounds per square inch (psi).
terms continue
to decrease;
they do Notation
not approach a finite number. Therefore, the sequence is divergent.
10-1 These
Sequences,
Series,
and Sigma
The sum of the first five terms is (–1) + (–2) + (–3) + (–4) + (–5) or –15.
65. SOLUTION: Find the first five terms of the sequence using the explicit formula,
.
These terms approach a finite number, 0. Therefore, the sequence is convergent.
The approximate sum of the first five terms is (–0.61) + (0.37) + (–0.22) + (0.14) + (–0.08) or about –0.4.
66. WATER PRESSURE The pressure exerted on the human body at sea level is 14.7 pounds per square inch (psi).
For each additional foot below sea level, the pressure is about 0.445 psi greater, as shown.
a. Write a recursive formula to represent a n, the pressure at n feet below sea level. (Hint: Let a 0 = 14.7.)
b. Write the first three terms of the sequence and describe what they represent.
c. Scuba divers cannot safely dive deeper than 100 feet. Write an explicit formula to represent a n. Then use the
formula to find the water pressure at 100 feet below sea level.
SOLUTION: a. Let a 0 = 14.7. n = 1 represents the human body 1 foot below sea level. Since the pressure exerted on the human
body increases about 0.445 psi for each foot below sea level, a 1 = 14.7 + 0.445 or 15.145 psi. When the human
body is 2 feet below sea level, n = 2 and a 2 = 15.145 + 0.445 or 15.59 psi. a 1 and a 2 can be written as:
a 1 = 14.7 + 0.445 = a 0 + 0.445
a 2 = 15.145 + 0.445 = a 1 + 0.445
So, a recursive formula to represent the pressure at n feel below sea level is a 0 = 14.7, a n = a n – 1 + 0.445.
b. From part a, a 0 = 14.7, a 1 = 15.145, and a 2 = 15.59. a 0 represents the pressure exerted on the human body at
sea level. The pressure exerted on the human body at sea level is 14.7 psi. a 1 represents the pressure exerted on
the human body 1 foot below sea level. The pressure exerted on the human body 1 foot below sea level is 15.145
psi. a 2 represents the pressure exerted on the human body 2 feet below sea level. The pressure exerted on the
human body 2 feet below sea level is 15.59 psi.
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eSolutions
Page 38
c. The common difference between the terms is 0.445. This sequence can be represented by a linear function that
has slope 0.445. Therefore, a n = 0.445n + b, where b is a constant. Since a n = 14.7 when n = 0, substitute these
terms approach
a finite
0. Therefore,
10-1 These
Sequences,
Series,
andnumber,
Sigma
Notationthe sequence is convergent.
The approximate sum of the first five terms is (–0.61) + (0.37) + (–0.22) + (0.14) + (–0.08) or about –0.4.
66. WATER PRESSURE The pressure exerted on the human body at sea level is 14.7 pounds per square inch (psi).
For each additional foot below sea level, the pressure is about 0.445 psi greater, as shown.
a. Write a recursive formula to represent a n, the pressure at n feet below sea level. (Hint: Let a 0 = 14.7.)
b. Write the first three terms of the sequence and describe what they represent.
c. Scuba divers cannot safely dive deeper than 100 feet. Write an explicit formula to represent a n. Then use the
formula to find the water pressure at 100 feet below sea level.
SOLUTION: a. Let a 0 = 14.7. n = 1 represents the human body 1 foot below sea level. Since the pressure exerted on the human
body increases about 0.445 psi for each foot below sea level, a 1 = 14.7 + 0.445 or 15.145 psi. When the human
body is 2 feet below sea level, n = 2 and a 2 = 15.145 + 0.445 or 15.59 psi. a 1 and a 2 can be written as:
a 1 = 14.7 + 0.445 = a 0 + 0.445
a 2 = 15.145 + 0.445 = a 1 + 0.445
So, a recursive formula to represent the pressure at n feel below sea level is a 0 = 14.7, a n = a n – 1 + 0.445.
b. From part a, a 0 = 14.7, a 1 = 15.145, and a 2 = 15.59. a 0 represents the pressure exerted on the human body at
sea level. The pressure exerted on the human body at sea level is 14.7 psi. a 1 represents the pressure exerted on
the human body 1 foot below sea level. The pressure exerted on the human body 1 foot below sea level is 15.145
psi. a 2 represents the pressure exerted on the human body 2 feet below sea level. The pressure exerted on the
human body 2 feet below sea level is 15.59 psi.
c. The common difference between the terms is 0.445. This sequence can be represented by a linear function that
has slope 0.445. Therefore, a n = 0.445n + b, where b is a constant. Since a n = 14.7 when n = 0, substitute these
values into the equation and solve for b.
Thus, an explicit formula for a n is a n = 0.445n + 14.7. Substitute n = 100 into this formula to find the water
pressure at 100 feet below sea level.
Therefore, the water pressure at 100 feet below sea level is 59.2 psi.
Match each sequence with its graph.
a. a =
n
n
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b. a n = –
n+9
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10-1 Sequences, Series, and Sigma Notation
Therefore, the water pressure at 100 feet below sea level is 59.2 psi.
Match each sequence with its graph.
a. a n =
n
b. a n = –
n+9
c.
d. a n = 8 –
(2n)
e. a n = 9 – 2n
f.
67. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
, the first three terms are
,
, and
. So, three points are
is the only sequence that has points that match this graph. Therefore, the correct answer is b.
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Page 40
10-1 Sequences, is the only sequence that has points that match this graph. Therefore, the correct answer is b.
Series, and Sigma Notation
68. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
, the first three terms are
,
, and
. So, three points are
is the only sequence that has points that match this graph. Therefore, the correct answer is f.
69. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
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eSolutions
For
, the first three terms are
Page 41
.So, three points are
10-1 Sequences, is the only sequence that has points that match this graph. Therefore, the correct answer is f.
Series, and Sigma Notation
69. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
, the first three terms are
,
, and
. So, three points are
is the only sequence that has points that match this graph. Therefore, the correct answer is c.
70. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
eSolutions
Manual - Powered by Cognero
For
, the first three terms are
Page 42
.So, three points are
10-1 Sequences,
Series, and Sigma Notation
is the only sequence that has points that match this graph. Therefore, the correct answer is c.
70. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
an =
, the first three terms are
,
, and
. So, three points are
n is the only sequence that has points that match this graph. Therefore, the correct answer is a.
71. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
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eSolutions
For
, the first three terms are
Page 43
.So, three points are
10-1 aSequences,
Series,
andthat
Sigma
Notation
n is the only
sequence
has points
that match this graph. Therefore, the correct answer is a.
n=
71. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
, the first three terms are
,
, and
. So, three points are
a n = 9 − 2n is the only sequence that has points that match this graph. Therefore, the correct answer is e.
72. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
eSolutions Manual - Powered by Cognero
For
, the first three terms are
Page 44
.So, three points are
10-1 aSequences,
Series, and Sigma Notation
= 9 − 2n is the only sequence that has points that match this graph. Therefore, the correct answer is e.
n
72. SOLUTION: Find the first few terms of each sequence and identify the point (n, a n) that corresponds to each term.
For a n =
n, the first three terms are
,
, and 4. So, three points are
For
, the first three terms are 8.25, 7.5, and 6.75. So, three points are (1, 8.25), (2, 7.5), and (3, 6.75).
For
, the first three terms are
.So, three points are
For
, the first three terms are 6.5, 5, and 2. So, three points are (1, 6.5), (2, 5), and (3, 2).
For a n = 9 − 2n, the first three terms are 7, 5, and 3. So, three points are (1, 7), (2, 5), and (3, 3).
For
, the first three terms are
,
, and
. So, three points are
is the only sequence that has points that match this graph. Therefore, the correct answer is d.
73. GOLDEN RATIO Consider the Fibonacci sequence 1, 1, 2, 3, …, a n – 2 + a n – 1.
a. Find
for the second through eleventh terms of the Fibonacci sequence.
b. Sketch a graph of the terms found in part a. Let n − 1 be the x-coordinate and
be the y-coordinate.
c. Based on the graph found in part b, does this sequence appear to be convergent? If so, describe the limit to three
decimal places. If not, explain why not.
d. In a golden rectangle, the ratio of the length to the width is about 1.61803399. This is called the golden ratio.
How does the limit of the sequence
compare to the golden ratio?
e . Golden rectangles are common in art and architecture. The Parthenon, in Greece, is an example of how golden
rectangles are used in architecture.
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Page 45
decimal places. If not, explain why not.
d. In a golden rectangle, the ratio of the length to the width is about 1.61803399. This is called the golden ratio.
How does the limit of the sequence
compare to the golden ratio?
10-1 eSequences,
Series, and Sigma Notation
.
Golden rectangles are common in art and architecture. The Parthenon, in Greece, is an example of how golden
rectangles are used in architecture.
Research golden rectangles and find two more examples of golden rectangles in art or architecture.
SOLUTION: a. The first 11 terms of the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, and 89. Find
for the second term.
Find
for the third term.
Find
for the fourth term.
Find
for the fifth term.
Find
for the sixth term.
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Page 46
10-1 Sequences, Series, and Sigma Notation
Find
for the sixth term.
Find
for the seventh term.
Find
for the eighth term.
Find
for the ninth term.
Find
for the tenth term.
Find
for the eleventh term.
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10-1 Sequences, Series, and Sigma Notation
Find
for the eleventh term.
for the second through eleventh terms of the Fibonacci sequence are ,
,
,
,
,
,
,
,
,
.
b. Graph the points
.
c. yes; As n increases,
approaches 1.618. Therefore, the sequence appears to be convergent.
d. Since the golden ratio is 1.61803399 and the limit of the sequence found in part c is 1.618, the two ratios are
equivalent to three decimal places.
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Page 48
e . Sample answer: The golden rectangle can be found throughout some of Leonardo Da Vinci’s art, including the
Mona Lisa. It was also used in the construction of the Taj Mahal in India.
c. yes; As n increases,
approaches 1.618. Therefore, the sequence appears to be convergent.
10-1 Sequences, Series, and Sigma Notation
d. Since the golden ratio is 1.61803399 and the limit of the sequence found in part c is 1.618, the two ratios are
equivalent to three decimal places.
e . Sample answer: The golden rectangle can be found throughout some of Leonardo Da Vinci’s art, including the
Mona Lisa. It was also used in the construction of the Taj Mahal in India.
Determine whether each sequence is convergent or divergent.
74. SOLUTION: The terms of the sequence appear to be increasing by a constant rate. Therefore, the terms of the sequence do not
approach a unique number and the sequence is divergent.
75. SOLUTION: The terms of the sequence appear to alternate signs while their absolute values seem to increase without bound.
Therefore, the terms of the sequence do not approach a unique number and the sequence is divergent.
76. SOLUTION: The terms of the sequence appear to alternate signs while slowly approaching 0. Therefore, the terms of the
sequence approach a unique number, 0, and the sequence is convergent.
77. eSolutions Manual - Powered by Cognero
SOLUTION: Page 49
The terms of the sequence appear to be approaching a value of about 2.75. Therefore, the terms of the sequence
76. SOLUTION: terms of theSeries,
sequence and
appearSigma
to alternate
signs while slowly approaching 0. Therefore, the terms of the
10-1 The
Sequences,
Notation
sequence approach a unique number, 0, and the sequence is convergent.
77. SOLUTION: The terms of the sequence appear to be approaching a value of about 2.75. Therefore, the terms of the sequence
approach a unique number, about 2.75, and the sequence is convergent.
Write an explicit formula for each recursively defined sequence.
78. a 1 = 10; a n = a n − 1 + 5
SOLUTION: Find the next two terms of the sequence.
n =2
n =3
Each term of the sequence is found by adding 5 to the succeeding term. Thus, the terms of the sequence are 10,
15, 20, 25, 30, … . The common difference between the terms is 5. This sequence can be represented by a linear
function that has slope 5. Therefore, a n = 5n + b, where b is a constant. Since a n = 10 when n = 1, substitute these
values into the equation and solve for b.
Thus, an explicit formula for the recursively defined sequence is a n = 5n + 5.
79. a 1 = 1.25; a n = a n − 1 − 0.5
SOLUTION: Find the next two terms of the sequence.
n =2
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Page 50
an explicitSeries,
formula for
the Sigma
recursively
defined sequence is a n = 5n + 5.
10-1 Thus,
Sequences,
and
Notation
79. a 1 = 1.25; a n = a n − 1 − 0.5
SOLUTION: Find the next two terms of the sequence.
n =2
n =3
Each term of the sequence is found by subtracting 0.5 to the succeeding term. Thus, the terms of the sequence are
1.25, 0.75, 0.25, −0.25, −0.75, … . The common difference between the terms is −0.5. This sequence can be
represented by a linear function that has slope −0.5. Therefore, a n = −0.5n + b, where b is a constant. Since a n =
1.25 when n = 1, substitute these values into the equation and solve for b.
Thus, an explicit formula for the recursively defined sequence is a n = −0.5n + 1.75.
80. a 1 = 128; a n = 0.5a n − 1
SOLUTION: Find the next two terms of the sequence.
n =2
n =3
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Each term of the sequence is found by multiplying the succeeding term by 0.5. Thus, the terms of the sequence are
128, 64, 32, 16, 8, 4, … . The terms of the sequence can be written in terms of multiples of 0.5.
0
1−1
10-1 Thus,
Sequences,
Series, and Sigma Notation
an explicit formula for the recursively defined sequence is a
n=
−0.5n + 1.75.
80. a 1 = 128; a n = 0.5a n − 1
SOLUTION: Find the next two terms of the sequence.
n =2
n =3
Each term of the sequence is found by multiplying the succeeding term by 0.5. Thus, the terms of the sequence are
128, 64, 32, 16, 8, 4, … . The terms of the sequence can be written in terms of multiples of 0.5.
0
1−1
a 1 = 128(0.5) or 128(0.5)
1
2−1
2
3−1
3
4−1
4
5−1
a 2 = 128(0.5) or 128(0.5)
a 3 = 128(0.5) or 128(0.5)
a 4 = 128(0.5) or 128(0.5)
a 5 = 128(0.5) or 128(0.5)
Thus, an explicit formula for the recursively defined sequence is a n = 128(0.5)
n−1
.
81. MULTIPLE REPRESENTATIONS In this problem, you will investigate sums of infinite series.
a. NUMERICAL Calculate the first five terms of the infinite sequence
b. GRAPHICAL Use a graphing calculator to sketch .
.
c. VERBAL Describe what is happening to the terms of the sequence as n
.
d. NUMERICAL Find the sum of the first 5 terms, 7 terms, and 9 terms of the series.
e . VERBAL Describe what is happening to the partial sums S n as n increases.
f. VERBAL Predict the sum of the first n terms of the series. Explain your reasoning.
SOLUTION: a. Substitute n = 1, 2, 3, 4, and 5 into
.
n =1
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n =2
Page 52
f. VERBAL Predict the sum of the first n terms of the series. Explain your reasoning.
SOLUTION: Substitute n = Series,
1, 2, 3, 4, and
.
10-1 a.Sequences,
and5 into
Sigma Notation
n =1
n =2
n =3
n =4
n =5
b. Plot the points (1, 0.4), (2, 0.04), (3, 0.004), (4, 0.0004), (5, 0.00004). Use a graphing calculator to find a few
other points and sketch the graph.
c. As n approaches infinity, the terms approach a value of 0. It seems as if each term adds another 0 to the right of
the decimal point in front of the 4.
d. To find S 5, add the first five terms. So, S 5 = 0.4 + 0.04 + 0.004 + 0.0004 + 0.00004 = 0.44444.
To find S 7, first find the sixth and seventh terms of the sequence.
n =6
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n =7
Page 53
d. To find S 5, add the first five terms. So, S 5 = 0.4 + 0.04 + 0.004 + 0.0004 + 0.00004 = 0.44444.
To find S , first find the sixth and seventh terms of the sequence.
7
10-1 Sequences,
Series, and Sigma Notation
n =6
n =7
Now add the sixth and seventh terms to S 5. So, S 5 = 0.44444 + 0.000004 + 0.0000004 = 0.4444444.
To find S 9, first find the eighth and ninth terms of the sequence.
n =8
n =9
Now add the eighth and ninth terms to S 7. So, S 9 = 0.4444444 + 0.00000004 + 0.000000004 = 0.444444444.
e . Sample answer: The sum seems to approach
, which is equivalent to
. So, the sum approaches
.
f. Sample answer: The sum will be the decimal 0.44…, where the number of 4s after the decimal point is n. The
sum of the first five terms has 5 fours, the sum of the first seven terms has 7 fours, and sum of the first nine terms
has 9 fours. Therefore, the sum of the first n terms will have n fours.
82. CHALLENGE Consider to the recursive sequence below.
for a 1 = 1, a 2 = 1, n
3
a. Find the first eight terms of the sequence.
b. Describe the similarities and differences between this sequence and the other recursive sequences in this lesson.
SOLUTION: a. The first two terms of the sequence are given as a 1 = 1 and a 2 = 1. Substitute n = 3, 4, 5, 6, 7, and 8 into the
recursive sequence to find the next 6 terms. eSolutions Manual - Powered by Cognero
Page 54
a. Find the first eight terms of the sequence.
b. Describe the similarities and differences between this sequence and the other recursive sequences in this lesson.
10-1 SOLUTION: Sequences, Series, and Sigma Notation
a. The first two terms of the sequence are given as a 1 = 1 and a 2 = 1. Substitute n = 3, 4, 5, 6, 7, and 8 into the
recursive sequence to find the next 6 terms. eSolutions Manual - Powered by Cognero
Page 55
10-1 Sequences, Series, and Sigma Notation
The first eight terms of the sequence are 1, 1, 2, 3, 3, 4, 5, and 5.
b. Sample answer: Like other recursive sequences, the terms in this sequence depend on previous terms. For
instance, after simplifying, a 4 = a 2 + a 3. However, this sequence is different because the previous terms are not
necessarily the two terms that came right before it. For instance, after simplifying, a 3 = a 2 + a 2.
83. OPEN ENDED Write a sequence either recursively or explicitly that has the following characteristics.
a. converges to 0
b. converges to 3
c. diverges
SOLUTION: a. Sample answer: Find a function f (x) such that
1,
,
,
, and
. The 100th term is
. So,
. Consider f (x) =
. Since
. The first five terms of f (x) are
, the sequence a n =
converges
to 0 as n approaches infinity.
b. Sample answer: For the sequence a n =
, as n approaches infinity, each term gets closer and closer to 0. Thus,
it converges to 0. If 3 is added to each term in the sequence, as n approaches infinity, the sequence will get closer
and closer to 3. Thus, the sequence a n =
+ 3 converges to 3.
c. Sample answer: For a sequence to diverge, it cannot have a limit. Linear functions do not have a limit because
they have a constant slope. Thus, the function f (x) = 2x does not have a limit. So, the sequence a n = 2n diverges.
84. WRITING IN MATH Describe why an infinite sequence must not only converge, but converge to 0, in order for
there to be a sum.
SOLUTION: Sample answer: If an infinite sequence converges to a value other than 0, then as the number of terms n
approaches infinity, the sum of the corresponding series S n will approach positive or negative infinity. Consider the sequence
. The first 5 terms of the sequence are 2.2, 2.02, 2.002, 2.0002, and 2.00002. Thus, the sequence converges to 2. S 1 = 2.2, eSolutions Manual - Powered by Cognero
S 2 = 2.2 + 2.02 = 4.22
S 3 = 2.2 + 2.02 + 2.002 = 6.22
Page 56
and closer to 3. Thus, the sequence a n =
+ 3 converges to 3.
c. Sample answer: For a sequence to diverge, it cannot have a limit. Linear functions do not have a limit because
10-1 they
Sequences,
Series, and Sigma Notation
have a constant slope. Thus, the function f (x) = 2x does not have a limit. So, the sequence a n = 2n diverges.
84. WRITING IN MATH Describe why an infinite sequence must not only converge, but converge to 0, in order for
there to be a sum.
SOLUTION: Sample answer: If an infinite sequence converges to a value other than 0, then as the number of terms n
approaches infinity, the sum of the corresponding series S n will approach positive or negative infinity. Consider the sequence
. The first 5 terms of the sequence are 2.2, 2.02, 2.002, 2.0002, and 2.00002. Thus, the sequence converges to 2. S 1 = 2.2, S 2 = 2.2 + 2.02 = 4.22
S 3 = 2.2 + 2.02 + 2.002 = 6.22
S 4 = 2.2 + 2.02 + 2.002 + 2.0002 = 8.2222
Each succeeding term increases the sum by about 2. Therefore, as n approaches infinity, the sum of the
corresponding series S n approaches infinity.
REASONING Determine whether each statement is true or false . Explain your reasoning.
85. SOLUTION: Find the sum for the expression on the left side,
.
So,
Find the sum for the expression on the right side,
. First find
.
So,
Now find
.
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S 4 = 2.2 + 2.02 + 2.002 + 2.0002 = 8.2222
Each succeeding term increases the sum by about 2. Therefore, as n approaches infinity, the sum of the
10-1 corresponding
Sequences,series
Series,
and Sigma Notation
S n approaches infinity.
REASONING Determine whether each statement is true or false . Explain your reasoning.
85. SOLUTION: Find the sum for the expression on the left side,
.
So,
Find the sum for the expression on the right side,
. First find
.
So,
Now find
.
So,
Thus,
Therefore, 86. SOLUTION: Find the sum for the expression on the left side,
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So,
.
Page 58
So,
Therefore, Thus,
10-1 Sequences, Series, and Sigma Notation
86. SOLUTION: Find the sum for the expression on the left side,
.
So,
Find the sum for the expression on the right side,
So,
Therefore, .
.
87. CHALLENGE Find the sum of the first 60 terms of the sequence below. Explain how you determined your
answer.
15, 17, 2, –15, –17, …,
where a n = a n – 1 – a n – 2 for n
3
SOLUTION: Write out the next several terms of the sequence.
Since a 7 = a 1 and a 8 = a 2, the sequence will repeat the first six terms. The sum of every 6 consecutive terms is 0.
So, the sum of 60 terms is 10 ⋅ 0 or 0.
88. WRITING IN MATH Write an outline that could be used to describe the steps involved in finding the 300th
partial sum of the infinite sequence a n = 2n − 3. Then explain how to express the same sum using sigma notation.
SOLUTION: Sample answer:
1. Calculate each term of the sequence.
a) Substitute n = 1 up to n = 300 into a n = 2n - 3 to find the value of each term in the sequence.
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b) Or, use a spreadsheet to automate the substitution of the 300 terms.
2. Sum all of the terms.
3. State the partial sum using the correct symbol and form.
Page 59
Since a = a and a = a , the sequence will repeat the first six terms. The sum of every 6 consecutive terms is 0.
7
1 Series,
8
2 and Sigma Notation
10-1 Sequences,
So, the sum of 60 terms is 10 ⋅ 0 or 0.
88. WRITING IN MATH Write an outline that could be used to describe the steps involved in finding the 300th
partial sum of the infinite sequence a n = 2n − 3. Then explain how to express the same sum using sigma notation.
SOLUTION: Sample answer:
1. Calculate each term of the sequence.
a) Substitute n = 1 up to n = 300 into a n = 2n - 3 to find the value of each term in the sequence.
b) Or, use a spreadsheet to automate the substitution of the 300 terms.
2. Sum all of the terms.
3. State the partial sum using the correct symbol and form.
a) The three hundredth partial sum is S 300 = 89,400.
4. Use the Greek letter sigma Σ to express the sum of the series.
a) Write the first term at the bottom of the sigma symbol Σ, n = 1 (called the index of summation).
b) Write the ending term at the top of the sigma symbol Σ, 300 (called the upper bound of summation).
c) State the sum using the correct symbol and form.
= 89,400.
Graph each complex number on a polar grid. Then express it in rectangular form.
89. SOLUTION: The value of r is 2, and the value of θ is
. Plot the polar coordinates
.
To express the number in rectangular form, evaluate the trigonometric values and simplify.
The rectangular form of
is
.
90. 2.5(cos 1 + i sin 1)
SOLUTION: eSolutions
Manual
by Cognero
The
value- Powered
of r is 2.5,
and the
value of θ is 1. Plot the polar coordinates
.
Page 60
rectangular form of
is
10-1 The
Sequences,
Series, and Sigma Notation
.
90. 2.5(cos 1 + i sin 1)
SOLUTION: The value of r is 2.5, and the value of θ is 1. Plot the polar coordinates
.
To express the number in rectangular form, evaluate the trigonometric values and simplify.
The rectangular form of 2.5(cos 1 + i sin 1)
is 1.35 + 2.10i
.
91. 5(cos 0 + i sin 0)
SOLUTION: The value of r is 5, and the value of θ is 0. Plot the polar coordinates
.
To express the number in rectangular form, evaluate the trigonometric values and simplify.
The rectangular form of 5(cos 0 + i sin 0) is 5.
Determine the eccentricity, type of conic, and equation of the directrix given by each polar equation.
92. eSolutions Manual - Powered by Cognero
Page 61
SOLUTION: Write the equation in standard form, r = .
To express the number in rectangular form, evaluate the trigonometric values and simplify.
10-1 Sequences, Series, and Sigma Notation
The rectangular form of 5(cos 0 + i sin 0) is 5.
Determine the eccentricity, type of conic, and equation of the directrix given by each polar equation.
92. SOLUTION: Write the equation in standard form, r = .
Since e = 0.25, the conic is a ellipse. For a polar equation of this form (where sinθ is included), the equation of the
directrix is x = d. From the numerator, we know that ed = 9, so d = 6. Therefore, the equation of the directrix is
x = 6.
93. SOLUTION: Write the equation in standard form, r =
.
Since e = 4, the conic is a hyperbola. For a polar equation of this form (where sinθ is included), the equation of the
directrix is y = d. From the numerator, we know that ed = 20, so d = 5. Therefore, the equation of the directrix is
y =5.
94. SOLUTION: Write the equation in standard form, r =
.
Since e = 1, the conic is a parabola. For a polar equation of this form (where sinθ is included), the equation of the
we know that ed =5, so d = 5. Therefore, the equation of the directrix isPage
y =62
5.
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directrix
x = d. From
the numerator,
Determine whether the points are collinear. Write yes or no.
Since e = 4, the conic is a hyperbola. For a polar equation of this form (where sinθ is included), the equation of the
is y = Series,
d. From the
numerator,
know that ed = 20, so d = 5. Therefore, the equation of the directrix is
10-1 directrix
Sequences,
and
Sigmawe
Notation
y =5.
94. SOLUTION: Write the equation in standard form, r =
.
Since e = 1, the conic is a parabola. For a polar equation of this form (where sinθ is included), the equation of the
directrix is x = d. From the numerator, we know that ed =5, so d = 5. Therefore, the equation of the directrix is y =
5.
Determine whether the points are collinear. Write yes or no.
95. (–3, –1, 4), (3, 8, 1), (5, 12, 0)
SOLUTION: Let a = (–3, –1, 4), b = (3, 8, 1), and c = (5, 12, 0). Form two vectors,
collinear, then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are
Use the formula for the angle between two vectors.
The angle between the vectors is about 69°. Therefore, the points are not collinear.
96. (4, 8, 6), (0, 6, 12), (8, 10, 0)
SOLUTION: Let a = (4, 8, 6), b = (0, 6, 12), and c = (8, 10, 0). Form two vectors,
then
the angle
between
will be 0° or 180°.
eSolutions
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- Powered
by Cognero and Find the component form of each vector.
and . If the three points are collinear,
Page 63
10-1 The angle between the vectors is about 69°. Therefore, the points are not collinear.
Sequences, Series, and Sigma Notation
96. (4, 8, 6), (0, 6, 12), (8, 10, 0)
SOLUTION: Let a = (4, 8, 6), b = (0, 6, 12), and c = (8, 10, 0). Form two vectors,
then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are collinear,
Use the formula for the angle between two vectors.
The angle between the vectors is 0°. Therefore, the points are collinear.
97. (0, –4, 3), (8, –10, 5), (12, –13, 2)
SOLUTION: Let a = (0, –4, 3), b = (8, –10, 5), and c = (12, –13, 2). Form two vectors,
collinear, then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are
Use the formula for the angle between two vectors.
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10-1 Sequences, Series, and Sigma Notation
The angle between the vectors is 0°. Therefore, the points are collinear.
97. (0, –4, 3), (8, –10, 5), (12, –13, 2)
SOLUTION: Let a = (0, –4, 3), b = (8, –10, 5), and c = (12, –13, 2). Form two vectors,
collinear, then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are
Use the formula for the angle between two vectors.
The angle between the vectors is about 137°. Therefore, the points are not collinear.
98. (–7, 2, –1), (–9, 3, –4), (–5, 1, 2)
SOLUTION: Let a = (–7, 2, –1), b = (–9, 3, –4), and c = (–5, 1, 2). Form two vectors,
collinear, then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are
Use the formula for the angle between two vectors.
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10-1 Sequences, Series, and Sigma Notation
The angle between the vectors is about 137°. Therefore, the points are not collinear.
98. (–7, 2, –1), (–9, 3, –4), (–5, 1, 2)
SOLUTION: Let a = (–7, 2, –1), b = (–9, 3, –4), and c = (–5, 1, 2). Form two vectors,
collinear, then the angle between
and will be 0° or 180°.
Find the component form of each vector.
and . If the three points are
Use the formula for the angle between two vectors.
The angle between the vectors is about 0°. Therefore, the points are collinear.
Find the length and the midpoint of the segment with the given endpoints.
99. (2, −15, 12), (1, −11, 15)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
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10-1 Sequences, Series, and Sigma Notation
The angle between the vectors is about 0°. Therefore, the points are collinear.
Find the length and the midpoint of the segment with the given endpoints.
99. (2, −15, 12), (1, −11, 15)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
100. (−4, 2, 8), (9, 6, 0)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
101. (7, 1, 5), (−2, −5, −11)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
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10-1 Sequences, Series, and Sigma Notation
101. (7, 1, 5), (−2, −5, −11)
SOLUTION: Use the Distance Formula for points in space to find the length of the segment.
Use the Midpoint Formula for points in space to find the midpoint of the segment.
102. TIMING The path traced by the tip of the hour-hand of a clock can be modeled by a circle with parametric
equations x = 6 sin t and y = 6 cos t.
a. Find an interval for t in radians that can be used to describe the motion of the tip as it moves from 12 o’clock
noon to 12 o’clock noon the next day.
b. Simulate the motion described in part a by graphing the equation in parametric mode on a graphing calculator.
c. Write an equation in rectangular form that models the motion of the hour-hand. Find the radius of the circle
traced out by the hour-hand if x and y are given in inches.
SOLUTION: a. The tip of the hour-hand will complete one full rotation around the clock between 12 o’clock noon and midnight.
It will complete another full rotation around the clock between midnight and 12 o’clock noon the next day.
Therefore, the hour-hand will complete two full rotations around the clock from 12 o’clock noon to 12 o’clock noon
the next day. Since the parametric equations are written in terms of the trigonometric functions sine and cosine,
one full rotation will be completed every 2π, which is the period of these two functions. Therefore, two full
rotations will be completed after 4π. So, an interval for t in radians that can be used to describe the motion of the
tip is 0 ≤ t ≤ 4π.
b. Enter the parametric equations into a graphing calculator. Adjust the window so that 0 ≤ t ≤ 4π.
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Page 68
Therefore, the hour-hand will complete two full rotations around the clock from 12 o’clock noon to 12 o’clock noon
the next day. Since the parametric equations are written in terms of the trigonometric functions sine and cosine,
one full rotation will be completed every 2π, which is the period of these two functions. Therefore, two full
will be completed after 4π. So, an interval for t in radians that can be used to describe the motion of the
10-1 rotations
Sequences,
Series, and Sigma Notation
tip is 0 ≤ t ≤ 4π.
b. Enter the parametric equations into a graphing calculator. Adjust the window so that 0 ≤ t ≤ 4π.
c. Solve the equations for sin t and cos t. Then use a trigonometric identity.
2
2
A rectangular equation that models the motion of the hour-hand is x + y = 36. This equation is in standard form,
2
so r = 36 or r = 6. Thus, the radius of the circle traced out by the hour-hand is 6 inches.
Find the exact value of each expression.
103. tan
SOLUTION: Write
as the sum or difference of angle measures with tangents that you know.
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Page 69
rectangular equation that models the motion of the hour-hand is x
10-1 ASequences,
Series, and Sigma Notation
2
2
2
+ y = 36. This equation is in standard form,
so r = 36 or r = 6. Thus, the radius of the circle traced out by the hour-hand is 6 inches.
Find the exact value of each expression.
103. tan
SOLUTION: Write
as the sum or difference of angle measures with tangents that you know.
104. sin 75°
SOLUTION: Write 75° as the sum or difference of angle measures with sines that you know.
105. cos 165°
SOLUTION: Write 165° as the sum or difference of angle measures with cosines that you know.
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10-1 Sequences, Series, and Sigma Notation
105. cos 165°
SOLUTION: Write 165° as the sum or difference of angle measures with cosines that you know.
Find the partial fraction decomposition of each rational expression.
106. SOLUTION: Rewrite the expression so that the numerator is not of equal or higher degree than the denominator.
Thus,
can be written as .
Rewrite the expression as partial fractions with constant numerators, A and B, and denominators that are the linear
factors of the original denominator.
2
Multiply each side by the LCD, 2x –3x + 1.
Group the like terms.
Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the
coefficients of the x-terms on the left side of the equation must equal the coefficients of the x-terms on the right
side.
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Use any method to solve the new system.
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10-1 Sequences, Series, and Sigma Notation
Find the partial fraction decomposition of each rational expression.
106. SOLUTION: Rewrite the expression so that the numerator is not of equal or higher degree than the denominator.
Thus,
can be written as .
Rewrite the expression as partial fractions with constant numerators, A and B, and denominators that are the linear
factors of the original denominator.
2
Multiply each side by the LCD, 2x –3x + 1.
Group the like terms.
Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the
coefficients of the x-terms on the left side of the equation must equal the coefficients of the x-terms on the right
side.
Use any method to solve the new system.
Replace A and B with –6 and 5 in the partial fraction decomposition.
107. eSolutions Manual - Powered by Cognero
Page 72
SOLUTION: Rewrite the expression as partial fractions with constant numerators, A and B, and denominators that are the linear
10-1 Sequences, Series, and Sigma Notation
107. SOLUTION: Rewrite the expression as partial fractions with constant numerators, A and B, and denominators that are the linear
factors of the original denominator.
2
Multiply each side by the LCD, 2x + x.
Group the like terms.
Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the
coefficients of the x-terms on the left side of the equation must equal the coefficients of the x-terms on the right
side.
Use any method to solve the new system.
Replace A and B with 3 and −2 in the partial fraction decomposition.
108. SOLUTION: This rational expression is proper. The denominator has one linear factor and one irreducible quadratic factor.
Rewrite the expression as partial fractions. For the denominator that is a quadratic factor, use Ax + B in the
numerator.
3
Multiply each side by the LCD, x + x.
eSolutions
Manual
by Cognero
Group
the- Powered
like terms.
Page 73
Replace A and B with 3 and −2 in the partial fraction decomposition.
10-1 Sequences, Series, and Sigma Notation
108. SOLUTION: This rational expression is proper. The denominator has one linear factor and one irreducible quadratic factor.
Rewrite the expression as partial fractions. For the denominator that is a quadratic factor, use Ax + B in the
numerator.
3
Multiply each side by the LCD, x + x.
Group the like terms.
Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the
coefficients of the x-terms on the left side of the equation must equal the coefficients of the x-terms on the right
side.
Use any method to solve the new system.
Replace A, B, and C with 1, −1, and 1in the partial fraction decomposition.
109. SAT/ACT The first term in a sequence is −5, and each subsequent term is 6 more than the term that immediately
precedes it. What is the value of the 104th term?
A 607
B 613
C 618
D 619
E 615
SOLUTION: eSolutions
Manual
If the
first- Powered
term is by
and each
–5Cognero
successive term is 6 more than the preceding term, then a n = –5 + 6(n – 1).
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10-1 Sequences, Series, and Sigma Notation
109. SAT/ACT The first term in a sequence is −5, and each subsequent term is 6 more than the term that immediately
precedes it. What is the value of the 104th term?
A 607
B 613
C 618
D 619
E 615
SOLUTION: If the first term is –5 and each successive term is 6 more than the preceding term, then a n = –5 + 6(n – 1).
The correct answer is B.
110. REVIEW Find the exact value of cos 2
if
and 180° <
< 270°.
F
G
H
J
SOLUTION: The correct answer is J.
111. The first four terms of a sequence are 144, 72, 36, and 18. What is the tenth term in the sequence?
A0
B
C
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D
SOLUTION: Page 75
10-1 Sequences, Series, and Sigma Notation
The correct answer is J.
111. The first four terms of a sequence are 144, 72, 36, and 18. What is the tenth term in the sequence?
A0
B
C
D
SOLUTION: 144, 72, 36, 18, …
Each term is
of the term preceding it, so the nth term can be given by a n = 144
.
The correct answer is C.
112. REVIEW How many 5-inch cubes can be stacked inside a box that is 10 inches long, 15 inches wide, and 5 inches tall?
F5
G6
H 15
J 20
SOLUTION: If the box is 10 inches long by 15 inches wide, then 2 five-inch cubes can be placed lengthwise and 3 cubes can be
placed widthwise for a total of 2 × 3 or 6 cubes.
The box is 5 inches high so only one layer of cubes can be stacked in the box for a total of 6 × 1 or 6 total cubes.
The correct answer is G.
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