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CCAlg2 Unit 12 Sequence and Series

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Lesson 12.1: Intro to Sequence and Recursion
Learning Goals:
1) What is a sequence? What notation do we use?
2) What is a recursive sequence and how can we find the π‘›π‘‘β„Ž term of it?
Sequence: a list of numbers or objects in a specific order
Example 1: 3, 7, 11, 15, 19, 23, …
Example 2: 0, 2, 0, 2, 0, 2, …
Example 3:
Terms: each member of a sequence
3, 7, 11, 15, 19, 23, …
0, 2, 0, 2, 0, 2, …
1st term = 3
1st term = 0
1st term = 1
4th term = 15
4th term = 2
4th term = 7
1
Sequences
A sequence is a function whose domain is a set of positive integers. If a domain is
not specified, it is understood that the domain starts with 1. The values in the range
are called the terms of the sequence.
A finite sequence has a limited number of terms.
𝑛=1
2
3
4
An infinite sequence continues without stopping.
π‘Žπ‘› = 2
4
6
8
Finite sequence: 2, 4, 6, 8
Infinite sequence: 2, 4, 6, 8, …
A sequence can be specified by an equation, or rule. For example, both sequences
above can be described by the rule π‘Žπ‘› = 2𝑛 or 𝑓(𝑛) = 2𝑛
π‘Ž3 = 2(3) = 6
or 𝑓(4) = 2(4) = 8
Example 1: Write the first six terms of π‘Žπ‘› = 2𝑛+1 the formula is the rule for the
sequence!
𝑛=1
𝑛=2
𝑛=3
𝑛=4
𝑛=5
𝑛=6
21+1
22+1
23+1
24+1
25+1
26+1
π‘Ž2 = 8
π‘Ž3 = 16
π‘Ž4 = 32
π‘Ž1 = 4
π‘Ž5 = 64
π‘Ž6 = 128
{4, 8, 16, 32, 63, 128}
2
Example 2: Write the first 4 terms of the sequence: π‘Žπ‘› =
1+(−1)𝑛
𝑛
𝑛=1
𝑛=2
𝑛=3
𝑛=4
1+(−1)1
1+(−1)2
1+(−1)3
1+(−1)4
1
2
3
4
π‘Ž1 = 0
π‘Ž2 = 1
π‘Ž3 = 0
π‘Ž4 =
1
2
1
{0, 1, 0, }
2
Find the indicated term of the sequence:
3.
π‘Žπ‘› = (−1)𝑛 (3𝑛 − 2)
4. 𝑓 (𝑛) =
π‘Ž25 = (−1)25 (3(25) − 2)
𝑓(11) =
π‘Ž25 = (−1)(75 − 2)
𝑓(11) =
π‘Ž25 = −73
𝑓(11) =
4𝑛
2𝑛2 −3
4(11)
2(11)2 −3
44
2(121)−3
44
239
Writing the Rule for Sequences
3
Describe the pattern, write the next term, and write a rule for the 𝑛th term of the
sequence
{2,
𝑛=1
4,
6,
8, … }
2
3
4
{2,
𝑛=1
×2 ↓ ×2 ↓ ×2 ↓ ×2 ↓
π‘Žπ‘› = 2
4
6
3,
4,
5, … }
2
3
4
+1 ↓ +1 ↓ +1 ↓ +1 ↓
8
π‘Žπ‘› = 2
π‘Žπ‘› = 2𝑛 or 𝑓(𝑛) = 2𝑛
3
4
5
π‘Žπ‘› = 𝑛 + 1 or 𝑓(𝑛) = 𝑛 + 1
5. Describe the pattern, write the next term, and write a rule for the 𝑛th term of
the sequence.
a)
1,
𝑛=1
4,
9,
16, …
2
3
4
^2 ↓ ^2 ↓
π‘Žπ‘› = 1
^2 ↓ ^2 ↓
4
9
16
π‘Žπ‘› = 𝑛2 or 𝑓(𝑛) = 𝑛2
1 1
c) { ,
,
1
,
1
3 9 27 81
𝑛=1
π‘Žπ‘› =
2
1
3𝑛
b)
0,
7,
26,
63, …
𝑛=1
2
3
4
↓
↓
↓
↓
7
26
63
π‘Žπ‘› = 0
π‘Žπ‘› = 𝑛3 − 1or 𝑓(𝑛) = 𝑛3 − 1
,…}
3
4
or 𝑓 (𝑛) =
1
3𝑛
6. The diagrams below represent the first three terms of a sequence. Assuming
the pattern continues, which formula determines π‘Žπ‘› , the number of shaded
squares in the 𝑛th term?
(1) π‘Žπ‘› = 4𝑛 + 12
(2) π‘Žπ‘› = 4𝑛 + 8
(3) π‘Žπ‘› = 4𝑛 + 4
(4) π‘Žπ‘› = 4𝑛 + 2
4
Vocabulary:
Explicit Rule: a formula that defines a sequence
2, 4, 6, 8, … (sequence)
π‘Žπ‘› = 2𝑛 or 𝑓(𝑛) = 2𝑛 (explicit rule)
Recursive Rule: a formula that requires the computation of the previous term in
order to find the next term of a sequence. (1) starting term (2) rule that uses
previous term
The example below is a recursive sequence because you need the first term to
find the second term.
π‘Ž1 = −4
π‘Žπ‘› = π‘Žπ‘›−1 + 5
π‘Ž1 = −4
π‘Ž2 = π‘Ž2−1 + 5 = π‘Ž1 + 5 = −4 + 5 = 1
π‘Ž3 = π‘Ž3−1 + 5 = π‘Ž2 + 5 = 1 + 5 = 6
π‘Ž4 = π‘Ž4−1 + 5 = π‘Ž3 + 5 = 6 + 5 = 11
5
Finding Terms of Recursive Sequences
7. Write the first four terms of the recursive sequence given below:
π‘Ž1 = 2
π‘Žπ‘› = π‘Žπ‘›−1 − 3
π‘Ž1 = 2
π‘Ž2 = π‘Ž2−1 − 3 = π‘Ž1 − 3 = 2 − 3 = −1
π‘Ž3 = π‘Ž3−1 − 3 = π‘Ž2 − 3 = −1 − 3 = −4
π‘Ž4 = π‘Ž4−1 − 3 = π‘Ž3 − 3 = −4 − 3 = −7
8. Write the first five terms of the recursive sequence given below:
𝑓(1) = 4
𝑓(𝑛) = 2𝑓(𝑛 − 1)
𝑓(1) = 4
𝑓(2) = 2𝑓(2 − 1) = 2𝑓(1) = 2(4) = 8
𝑓(3) = 2𝑓(3 − 1) = 2𝑓(2) = 2(8) = 16
𝑓(4) = 2𝑓(4 − 1) = 2𝑓(3) = 2(16) = 32
𝑓(5) = 2𝑓(5 − 1) = 2𝑓(4) = 2(32) = 64
9. For the recursively defined sequence 𝑑𝑛 = (𝑑𝑛−1 )2 + 2 and 𝑑1 = 2 the value of
𝑑4 is
(1) 18
(2) 38
(3) 456
(4) 1446
𝑑1 = 2
𝑑2 = (𝑑2−1 )2 + 2 = (𝑑1 )2 + 2 = 22 + 2 = 4 + 2 = 6
𝑑3 = (𝑑3−1 )2 + 2 = (𝑑2 )2 + 2 = 62 + 2 = 36 + 2 = 38
𝑑4 = (𝑑4−1 )2 + 2 = (𝑑3 )2 + 2 = 382 + 2 = 1444 + 2 = 1446
6
Homework 12.1: Intro to Sequence and Recursion
1. Given the sequence 7, 11, 15, 19, …, which of the following represents a
formula that will generate it?
(1) π‘Ž(𝑛) = 4𝑛 + 7
(2) π‘Ž(𝑛) = 3𝑛 + 4
(3) π‘Ž(𝑛) = 3𝑛 + 7
(4) π‘Ž(𝑛) = 4𝑛 + 3
2. Determine and label the first four terms of each of the following sequences.
2
(a) π‘Žπ‘› = 𝑛 − 5
2 𝑛
(b) 𝑑(𝑛) = ( )
3
(c) 𝑑𝑛 =
1
𝑛+1
3. Which of the following formulas would represent the sequence
10, 20, 40, 80, 160, …
(1) π‘Žπ‘› = 10𝑛
(2) π‘Žπ‘› = 10(2)𝑛
(3) π‘Žπ‘› = 5(2)𝑛
(4) π‘Žπ‘› = 2𝑛 + 10
4. Write an algebraic expression that represents π‘Žπ‘› for each sequence.
(a) 3, 9, 27, 81, …
(b)
1 1 1 1
, , , ,…
2 3 4 5
(c) 1, √2, √3, 2, …
5. A recursive sequence is defined by π‘Žπ‘›+1 = 2π‘Žπ‘› − π‘Žπ‘›−1 with π‘Ž1 = 0 and π‘Ž2 = 1.
Which of the following represents the value of π‘Ž5 ? (1) 8 (2) −7 (3) 3 (4) 4
7
6. Sequences below are defined recursively. Determine and label the next three
terms of the sequence.
(a) π‘Ž1 = 5, π‘Žπ‘› = π‘Žπ‘›−1 + 1
(b) π‘Ž1 = −2, π‘Žπ‘› = −2π‘Žπ‘›−1
(c) 𝑏𝑛 = 𝑏𝑛−1 + 2𝑛 with 𝑏1 = 5
7. Sean has started an exercise program. The first day he worked out for 30
minutes. Each day for the next six days, he increased his time by 5 minutes.
(a) Write a sequence for the number of minutes that Sean worked out for each of
the seven days.
(b) Write a recursive definition for this sequence.
8. A pattern of blocks is shown below. If the pattern of blocks continues, which
formula(s) could be used to determine the number of blocks in the 𝑛th term?
I
π‘Žπ‘› = 𝑛 + 4
(1) I and II
(2) I and III
II
III
π‘Ž1 = 2
π‘Žπ‘› = 4𝑛 − 2
π‘Žπ‘› = π‘Žπ‘›−1 + 4
(3) II and III (4) III, only
8
12.2: Arithmetic and Geometric Sequences
Learning Goals:
1) What is an arithmetic sequence? What is the common difference of an
arithmetic sequence?
2) How can we find the 𝑛th term of an arithmetic sequence?
3) What is a geometric sequence? What is the common ratio of a geometric
sequence?
4) How can we find the 𝑛th term of a geometric sequence?
Arithmetic Sequence: a sequence where you are adding a fixed amount from
one term to the next
1, 3, 5, 7, 9, … adding 2
4, 2, 0, −2, −4 subtracting 2
Common Difference: (𝑑 = π‘Ž2 − π‘Ž1 ) the fixed amount added to get to the next
term.
1, 3, 5, 7, 9, … 𝑑 = 2
4, 2, 0, −2, −4 𝑑 = − 2
𝑑 = 3 − 1 = 2 or d = 9 − 7 = 2
𝑑 = 2 − 4 = −2 or d = −4 − (−2) = −2
Identify Arithmetic Sequences:
Example 1: Tell whether the sequence −5, −3, −1, 1, 3, … is arithmetic.
Find the differences of consecutive terms.
π‘Ž2 − π‘Ž1 = −3 − (−5) = 2
π‘Ž3 − π‘Ž2 = −1 − (−3) = 2
π‘Ž4 − π‘Ž3 = 1 − (−1) = 2
π‘Ž5 − π‘Ž4 = 3 − 1 = 2
Arithmetic because you add 2 to get the next term!
Arithmetic sequences can be represented by a linear
scatter plot.
𝐷 = −5, −3, −1, 1, 3, …
𝑅 = 2, 4, 6, 8, …
9
RULE FOR AN ARITHMETIC SEQUENCE
The 𝑛th term of an arithmetic sequence with the first term π‘Ž1 and common
difference 𝑑 is given by:
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑 On Reference Sheet!
Example 2: Write a rule for the 𝑛th term of the sequence. Then find π‘Ž19
a. 2, 9, 16, 23, …
π‘Ž1 = 2
Common Difference: 𝑑 = π‘Ž2 − π‘Ž1 = 9 − 2 = 7 or π‘Ž3 − π‘Ž2 = 16 − 9 = 7
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑
π‘Žπ‘› = 7𝑛 − 5
π‘Žπ‘› = 2 + (𝑛 − 1)7
π‘Ž19 = 7(19) − 5
π‘Žπ‘› = 2 + 7𝑛 − 7
π‘Ž19 = 128
π‘Žπ‘› = 7𝑛 − 5
b. 57, 45, 33, 21, …
π‘Ž1 = 57
Common Difference: 𝑑 = π‘Ž2 − π‘Ž1 = 45 − 57 = −12 or π‘Ž3 − π‘Ž2 = 33 − 45 = −12
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑
π‘Žπ‘› = −12𝑛 + 69
π‘Žπ‘› = 57 + (𝑛 − 1)(−12)
π‘Ž19 = −12(19) + 69
π‘Žπ‘› = 57 − 12𝑛 + 12
π‘Ž19 = −159
π‘Žπ‘› = −12𝑛 + 69
Example 3: One term of an arithmetic sequence is π‘Ž11 = 41. The common
difference is 𝑑 = 5. Write a rule for the 𝑛th term.
π‘Ž11 = 41 so 𝑛 = 11 𝑑 = 5
π‘Ž1 =?
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑
π‘Ž11 = π‘Ž1 + (11 − 1)5
π‘Žπ‘› = −9 + (𝑛 − 1)5
41 = π‘Ž1 + (10)5
or
π‘Žπ‘› = −9 + 5𝑛 − 5
41 = π‘Ž1 + 50
or
π‘Žπ‘› = 5𝑛 − 14
π‘Ž1 = −9
10
Geometric Sequence: a sequence where you are multiplying by a fixed amount
from one term to the next
Common Ratio: (π‘Ÿ =
π‘Ž2
π‘Ž1
6
54
2
18
1
3
)the fixed amount multiplied to get to the next term
18, 6, 2 …
2, 6, 18, 54, … r = 3
π‘Ÿ = = 3 or
multiplying by
18, 6, 2 …
2, 6, 18, 54, … multiplying by 3
=3
π‘Ÿ=
6
18
=
r=
1
3
or
1
3
2
6
=
1
3
Geometric because you multiply to get the next term!
Geometric sequences can be represented by an
exponential scatter plot.
𝐷 = 1, 2, 3, 4, …
𝑅 = 1, 2, 4, 8, …
Identify Geometric Sequences:
Example 1: Tell whether the sequence 1, −4, 16, −64, 256, … is geometric.
To decide whether a sequence is geometric, find the ratios of consecutive terms.
π‘Ž2
π‘Ž1
=
−4
1
= −4
π‘Ž3
π‘Ž2
=
16
−4
= −4
π‘Ž4
π‘Ž3
=
−64
16
= −4
π‘Ž5
π‘Ž4
=
256
−64
= −4
Geometric because you multiply by −4 to get to the next term.
RULE FOR A GEOMETRIC SEQUENCE
The 𝑛th term of a geometric sequence with the first term π‘Ž1 and common
ratio π‘Ÿ is given by:
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1 On Reference Sheet!
11
Example2: Write a rule for the 𝑛th term of the sequence 972, −324, 108, −36, …
Then find π‘Ž10 .
1
Geometric because you multiply by − .
π‘Ÿ=
3
−324
972
=
−36
108
=−
1
3
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
1 𝑛−1
π‘Žπ‘› = 972 (− )
3
1 10−1
π‘Ž10 = 972 (− )
3
1 9
4
3
81
= 972 (− ) = −
Example 3*: One term of a geometric sequence is π‘Ž3 = −18. The common ratio
is π‘Ÿ = 3. Write a rule for the 𝑛th term.
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
π‘Ž3 = π‘Ž1 (3)3−1
π‘Žπ‘› = (−2) (3)𝑛−1
−18 = π‘Ž1 (3)2
−18 = 9π‘Ž1
π‘Ž1 = −2
12
Homework 12.2: Arithmetic and Geometric Sequences
1. Does the following pair of formulas represent the same sequence? Justify
your reasoning.
π‘Žπ‘›+1
2
= π‘Žπ‘› ,
3
π‘Ž0 = −1,
2 𝑛
𝑏𝑛 = − ( ) ,
3
and
for 𝑛 ≥ 0
2. A sequence has the following terms: π‘Ž1 = 4, π‘Ž2 = 10, π‘Ž3 = 25, π‘Ž4 = 62.5.
Which formula represents the 𝑛th term in the sequence?
(1) π‘Žπ‘› = 4 + 2.5𝑛
(2) π‘Žπ‘› = 4 + 2.5(𝑛 − 1)
(3) π‘Žπ‘› = 4(2.5)𝑛
(4) π‘Žπ‘› = 4(2.5)𝑛−1
3. Which arithmetic sequence has a common difference of 4?
(1) {0, 4𝑛, 8𝑛, 12𝑛, … }
(2) {𝑛, 4𝑛, 16𝑛, 64𝑛, … }
(3) {𝑛 + 1, 𝑛 + 5, 𝑛 + 9, 𝑛 + 13, … }
(4) {𝑛 + 4, 𝑛 + 16, 𝑛 + 64, 𝑛 + 256, … . }
4. An arithmetic sequence has a first term of 10 and a sixth term of 40. What is
the 20th term of this sequence?
(1) 105 (2) 110 (3) 124 (4) 130
5. What is the fourth term of the sequence defined by π‘Ž1 = 3π‘₯𝑦 5 and
2π‘₯
(1) 12π‘₯ 3 𝑦 3
π‘Žπ‘› = ( ) π‘Žπ‘›−1 ?
𝑦
(2) 24π‘₯ 2 𝑦 4
6. What is the common ratio of the sequence
(1) −
3𝑏
2π‘Ž2
(2) −
6𝑏
π‘Ž2
(3) −
3π‘Ž2
𝑏
(4) −
1
64
(3) 24π‘₯ 4 𝑦 2
π‘Ž5 𝑏 3 , −
3
32
(4) 48π‘₯ 5 𝑦
π‘Ž3 𝑏 4 ,
9
16
π‘Žπ‘ 5 , …?
6π‘Ž2
𝑏
7. What is the common ratio of the geometric sequence whose first term is 27
and the fourth term is 64?
(1)
3
4
(2)
64
81
(3)
4
3
(4)
37
3
13
Lesson 12.3: Recursive Sequences vs. Explicit Form
Learning Goal: How do we write a recursive sequence in explicit form and vice
versa?
Write the first four terms of the recursive sequence given below:
π‘Ž1 = −1
π‘Žπ‘› = 2π‘Žπ‘›−1 + 𝑛
π‘Ž2 = 2π‘Ž2−1 + 2 = 2π‘Ž1 + 2 = 2(−1) + 2 = −2 + 2 = 0
π‘Ž3 = 2π‘Ž3−1 + 3 = 2π‘Ž2 + 3 = 2(0) + 3 = 0 + 3 = 3
π‘Ž4 = 2π‘Ž4−1 + 4 = 2π‘Ž3 + 4 = 2(3) + 4 = 6 + 4 = 10
Notation: Recursive forms work with the term(s)
immediately in front of the term being examined. The
table shows that there are many options as to how this
relationship may be expressed in notations.
A recursive formula is written with two parts: a
statement of the first term along with a statement of the
formula relating successive terms.
The statements below are all naming the same
sequence:
π‘Ž1 = 1;
π‘Ž1 = 1;
π‘Žπ‘› = π‘Žπ‘›−1 + 4
π‘Žπ‘›−1 = π‘Žπ‘› + 4
𝑓(1) = 1;
𝑓(𝑛) = 𝑓(𝑛 − 1) + 4
𝑓(1) = 1;
𝑓(𝑛 + 1) = 𝑓(𝑛) + 4
*For multiple choice questions, you need to know that all of
these mean the same thing!
14
Rewrite the given sequence 3 other ways, using the notation in the above
example.
π‘Ž1 = −6
π‘Ž1 = 2
π‘Žπ‘›+1 = π‘Žπ‘› + 5
π‘Žπ‘› = 3π‘Žπ‘›−1
𝑓(1) = −6
𝑓(1) = 2
𝑓(𝑛 + 1) = 𝑓(𝑛) + 5
𝑓(𝑛 + 1) = 3𝑓(𝑛)
𝑓(1) = −6
𝑓(1) = 2
𝑓(𝑛) = 𝑓(𝑛 − 1) + 5
𝑓(𝑛) = 3𝑓(𝑛 − 1)
π‘Ž1 = −6
π‘Ž1 = 2
π‘Žπ‘› = π‘Žπ‘›−1 + 5
π‘Žπ‘›+1 = 3π‘Žπ‘›
Arithmetic! Adding 5
Geometric! Multiply by 3
Converting from Recursive to Explicit
Use the following explicit formulas with π‘Ž1 as the first term (Use for 𝑛 ≥ 1)
ARITHMETIC SEQUENCE:
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑
GEOMETRIC SEQUENCE:
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
Example 1: Convert each of the following recursive formulas to explicit formulas.
Identify each sequence as arithmetic, geometric, or neither.
a. π‘Žπ‘› = 4π‘Žπ‘›−1 with π‘Ž1 = 12
Geometric because you multiply by 4 to get the next term!
π‘Ž1 = 12 & π‘Ÿ = 4
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1 = 12(4)𝑛−1
b. π‘Žπ‘› = 4.2 + π‘Žπ‘›−1 with π‘Ž1 = 12
Arithmetic because you add 4.2 to get the next term!
π‘Ž1 = 12 & 𝑑 = 4.2
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑 = 12 + (𝑛 − 1)(4.2)
15
c. π‘Žπ‘›+1 = √5 π‘Žπ‘› with π‘Ž1 = 2
Geometric because you multiply by √5 to get the next term!
π‘Ž1 = 2 & π‘Ÿ = √5
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1 = 2(√5)𝑛−1
d. π‘Žπ‘›+1 = √5 + π‘Žπ‘› with π‘Ž1 = 2
Arithmetic because you add √5 to get the next term!
π‘Ž1 = 2 & 𝑑 = √5
π‘Žπ‘› = π‘Ž1 + (𝑛 − 1)𝑑 = 2 + (𝑛 − 1)(√5)
Convert from Explicit to Recursive
Use the following recursive formulas (MEMORIZE)
ARITHMETIC SEQUENCE:
Subscript notation:
π‘Ž1 = π‘₯ and π‘Žπ‘› = π‘Žπ‘›−1 + 𝑑
GEOMETRIC SEQUENCE:
Subscript notation:
π‘Ž1 = π‘₯ and π‘Žπ‘› = π‘Žπ‘›−1 π‘Ÿ
Function notation:
𝑓(1) = π‘₯ and 𝑓(𝑛) = 𝑓(𝑛 − 1)𝑑
Function notation:
𝑓(1) = π‘₯ and 𝑓(𝑛) = π‘Ÿ βˆ™ 𝑓(𝑛 − 1)
Example2: Write each sequence in recursive form.
1
a. π‘Žπ‘› = (3𝑛 ) for 𝑛 ≥ 0 so use 𝑛 = 0 for the 1st term (π‘Ž0 )
5
1
1
5
5
1
3
5
5
1
9
5
5
π‘Ž0 = (30 ) =
π‘Ž1 = (31 ) =
π‘Ž2 = (32 ) =
Geometric because you multiply by 3 to get the next term! π‘Ž0 =
1
5
& π‘Ÿ=3
π‘Žπ‘› = 3π‘Žπ‘›−1
16
b. π‘Žπ‘› = 16 − 2𝑛 for 𝑛 ≥ 1 so use 𝑛 = 1 for the 1st term (π‘Ž1 )
π‘Ž1 = 16 − 2(1) = 14
π‘Ž2 = 16 − 2(2) = 12
π‘Ž3 = 16 − 2(3) = 10
Arithmetic because you add −2 to get the next term!
π‘Ž1 = 14 & 𝑑 = −2
π‘Žπ‘› = π‘Žπ‘›−1 − 2
1 𝑛
c*. π‘Žπ‘› = 16 ( )
2
for 𝑛 ≥ 1 so use 𝑛 = 1 for the 1st term (π‘Ž1 )
1 1
π‘Ž1 = 16 ( ) = 8
2
1 2
π‘Ž2 = 16 ( ) = 4
2
1 3
π‘Ž3 = 16 ( ) = 2
2
1
π‘Ž1 = 8 & π‘Ÿ =
Geometric because you multiply by to get the next term!
2
1
2
1
π‘Žπ‘› = π‘Žπ‘›−1
2
6
d*. π‘Žπ‘› = 71 − 𝑛 for 𝑛 ≥ 0 so use 𝑛 = 0 for the 1st term (π‘Ž0 )
7
6
497
7
7
π‘Ž0 = 71 − (0) = 71 =
6
491
7
7
6
485
7
7
π‘Ž1 = 71 − (1) =
π‘Ž2 = 71 − (2) =
6
Arithmetic because you add − to get the next term!
7
π‘Žπ‘› = π‘Žπ‘›−1 −
π‘Ž0 = 71 & 𝑑 = −
6
7
6
7
17
Example 3: At her job, Pat earns $25,000 the first year and receives a raise of
$1,000 each year. The explicit formula for the nth term of this sequence is
π‘Žπ‘› = 25,000 + (𝑛 − 1)1000. Write a rule to represent the equivalent recursive
formula of this sequence.
π‘Ž1 = 25000
π‘Žπ‘› = π‘Žπ‘›−1 + 1000
Example 4: The initial value of a specialty toy is $3,000, and its value each of the
following years is 20% less. Write an explicit formula and a recursive formula to
model this sequence.
20% less means they have 80% left
π‘Ž1 =3000
π‘Ÿ=0.80
explicit: π‘Žπ‘› = 3000(0.80)𝑛−1
Recursive:
π‘Ž1 =3000
π‘Žπ‘› =0.80(π‘Žπ‘›−1 )
Example 5: The last row of an auditorium has 2500 seats, and each row before it
has 80 fewer seats than the row behind it. Write an explicit formula and a
recursive formula to model the sequence.
explicit: π‘Žπ‘› = 2500 + (𝑛 − 1)(−80)
Recursive:
π‘Ž1 =2500
π‘Žπ‘› =π‘Žπ‘›−1 −80
18
Homework 12.3: Recursive Sequences vs. Explicit Form
1. Write each recursive sequence in explicit form. Identify each sequence as
arithmetic, geometric, or neither.
a. π‘Ž1 = 3, π‘Žπ‘›+1 = π‘Žπ‘› + 5
b. π‘Ž1 = √2, π‘Žπ‘› =
π‘Žπ‘›−1
√2
2. Which of the following formulas is equivalent to the geometric sequence
modeled by the equation π‘Žπ‘› = 5(2)𝑛−1 ?
(1) π‘Ž1 = 5,
π‘Žπ‘›+1 = 2π‘Žπ‘›
(3) π‘Ž1 = 10,
π‘Žπ‘›+1 = 2π‘Žπ‘›
(2) π‘Ž1 = 2,
π‘Žπ‘›+1 = 5π‘Žπ‘›
(4) π‘Ž1 = 5,
π‘Žπ‘›+1 = (π‘Žπ‘› )2
3. If a geometric sequence has π‘Ž1 = 256 and π‘Ž8 = 512, find the exact value of
the common ratio π‘Ÿ.
4. Given the sequence 7, 11, 15, 19, … write an explicit formula to represent this
sequence.
5. Which recursively defined function has a first term equal to 10 and a common
difference of 4?
(1) 𝑓(1) = 10,
𝑓(π‘₯) = 𝑓(π‘₯ − 1) + 4
(2) 𝑓(1) = 4,
𝑓(π‘₯) = 𝑓(π‘₯ − 1) + 10
(3) 𝑓(1) = 10,
𝑓(π‘₯) = 4𝑓(π‘₯ − 1)
(4) 𝑓(1) = 4,
𝑓(π‘₯) = 10𝑓(π‘₯ − 1)
19
6. The recursive formula, π‘Ž2 = 18.5, π‘Žπ‘› = π‘Žπ‘›−1 + 1.5 models the number of
people, in millions, that own a smartphone in the U.S. 𝑛 years after 2014.
a. Write an equivalent explicit formula for the situation.
b. Find the value of π‘Ž5 using both the recursive formula and the explicit formula.
Explain what π‘Ž5 means in the context of the problem.
7. In an arithmetic sequence of numbers π‘Ž1 = −4 and π‘Ž6 = 46. Find the value
of π‘Ž12 .
8. Given that π‘Ž1 = 5 and π‘Ž2 = 15 are the first two terms of a geometric
sequence, determine the values of π‘Ž3 and π‘Ž10 . Show the calculations that lead
to your answers.
9. The formula below can be used to model which scenario?
π‘Ž1 = 2000
π‘Žπ‘›+1 = π‘Žπ‘› + 50
(1)
(2)
(3)
(4)
The last row of a concert hall has 2000 seats and each row before has
50 less seats than the one after it.
The first row of a concert hall has 2000 seats and each row after has 50
more seats than the previous row.
A bank account starts with a deposit of $2000 and it grows each year by
50%.
A used car is worth $2000 to start and depreciates by 50% each
subsequent year.
20
Lesson 12.4: Summation Notation
Learning Goals:
1) What is summation notation? How can we evaluate the sum of a finite
series given in summation notation?
2) How can we write a series in summation notation?
Summation Notation
ο‚· The summation of a specific number of terms of a sequence (a series) can
also be represented in a compact form, called summation notation, or
sigma notation.
ο‚· The Greek capital letter sigma, ∑ , is used to indicate a sum.
ο‚· Given a sequence π‘Ž1 , π‘Ž2 , π‘Ž3 , π‘Ž4, … we can write the sum of the first 𝑛 terms
of the sequence using the expression:
𝑛
∑ π‘Žπ‘˜ = π‘Ž1 + π‘Ž2 + π‘Ž3 + β‹― + π‘Žπ‘›
π‘˜=1
ο‚· It is read, “The sum of π‘Žπ‘˜ from π‘˜ = 1 to π‘˜ = 𝑛.
ο‚· Example:
5
∑ 2𝑛 = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) = 2 + 4 + 6 + 8 + 10 = 30
𝑛=1
ο‚· To write the terms of the series, replace 𝑛 by the consecutive integers from
1 to 5, as shown above.
2nd → stat → math → sum(5) & 2nd → stat → ops → seq 2π‘₯, π‘₯, 1, 5
21
Example 1: Write the sum without using summation notation. You do not need
to find the sum.
πŸ“
πŸ”
∑π’Œ = 0+1+2+3+4+5
∑ 𝒋 𝟐 = 52 + 62 + 72
π’Œ=𝟎
𝒋=πŸ“
πŸ’
∑
π’Š=𝟐
𝟏 1 1 1
= + +
π’Š 2 3 4
𝑖 is not an imaginary unit in this example!
πŸ”
∑ 𝒋𝒙𝒋−𝟏 = (1π‘₯ 1−1 ) + (2π‘₯ 2−1 ) + (3π‘₯ 3−1 ) + (4π‘₯ 4−1 ) + (5π‘₯ 5−1 ) + (6π‘₯ 6−1 )
𝒋=𝟏
= 1 + 2π‘₯ + 3π‘₯ 2 + 4π‘₯ 3 + 5π‘₯ 4 + 6π‘₯ 5
Example 2: Write the sum without using summation notation, and find the sum.
4
∑ 2π‘š + π‘₯ π‘š
π‘š=0
= (2(0) + π‘₯ 0 ) + (2(1) + π‘₯ 1 ) + (2(2) + π‘₯ 2 ) + (2(3) + π‘₯ 3 ) + (2(4) + π‘₯ 4 )
(2 + π‘₯) + (4 + π‘₯ 2 ) + (6 + π‘₯ 3 ) + (8 + π‘₯ 4 )
= (0 + 1) +
= 21 + π‘₯ + π‘₯ 2 + π‘₯ 3 + π‘₯ 4
3
∑(−4)𝑗−2 = (−4)0−2 + (−4)1−2 + (−4)2−2 + (−4)3−2
𝑗=0
= (−4)−2 + (−4)−1 + (−4)0 + (−4)1 =
=
1
1
+
+1−4
(−4)2 (−4)
1
1
51
+
−3=−
16 −4
16
22
3
∑ 𝑝 βˆ™ 3𝑝 = 102
math → 0: summation
𝑝=1
3
3 ∑ 𝑝 βˆ™ 3𝑝 = 3(102) = 306 just multiply the answer by 3
What if …
𝑝=1
4
∑ log (
π‘˜=1
π‘˜
) = −.6987
π‘˜+1
4
What if …
3 + ∑ log (
π‘˜=1
π‘˜
) = 3 − .6987 = −2.3010 just add 3 to the final sum
π‘˜+1
Example 3: Write each sum using summation notation.
Strategies:
ο‚· See if sequence is arithmetic or geometric→if it is you can find the explicit
formula for the sequence
ο‚· If the sequence is not arithmetic or geometric, try to look for other patterns
ο‚· Look to see if numbers are “famous” numbers like perfect squares or cubes
ο‚· ∑π‘›π‘˜=1 π‘Žπ‘˜
2 + 4 + 8 + 16 + β‹― + 512
π‘Ž1 = 2 & π‘Ÿ = 2
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
512 = 2(2)𝑛−1
256 = 2𝑛−1
get like “bases”
∑9π‘˜=1 2(2)π‘˜−1
28 = 2𝑛−1
8=𝑛−1
𝑛=9
𝑛
Example 4: What is the value of 3 ∑6𝑛=2 ?
2
(1) 10
(2) 13
(3) 30
(4) 60
23
Homework 12.4: Summation Notation
1. Write the sum without using summation notation, and find the sum.
a.
3
∑
𝑗=0
105
2𝑗 + 1
b.
5
∑(−1)π‘˜ π‘₯ π‘˜
π‘˜=0
c.
4
3 π‘š
∑ 16 ( )
2
π‘š=1
2. Which of the following represents the sum of a geometric sequence?
(1) ∑5𝑗=1(2𝑗 + 2𝑗 )
(2) ∑6π‘˜=2 2(3π‘˜−1 )
(3) ∑6𝑝=3(3𝑝 + 1)
(4) ∑6π‘š=2(2π‘š2 )
3. Write each sum using summation notation.
a. 2 + 4 + 6 + 8 + β‹― + 100
b. 12 + 22 + 32 + 42 + β‹― + 100002
c.
1
1βˆ™2
+
1
2βˆ™3
+
1
3βˆ™4
+ β‹―+
1
49βˆ™50
24
Lesson 12.5: Geometric Series
Learning Goal: What is a geometric series and how do we evaluate it?
Do Now: Answer the following questions in order to prepare for today’s lesson.
Geometric Sequence: 𝒂𝒏 = π’‚πŸ βˆ™ 𝒓𝒏−𝟏
1. Given the geometric sequence: −375, −75, −15, −3, …
a. Write the formula for the 𝑛th term of the geometric sequence.
π‘Ÿ=
−75
−375
=
1 𝑛−1
1
π‘Žπ‘› = −375 ( )
5
5
b. Use the formula to find the 7th term of the sequence. 𝑛 = 7
1 7−1
π‘Ž7 = −375 ( )
5
= −.024 = −
3
125
c. Find the sum of the first 7 terms of this sequence.
3
3
5
25
−375, −75, −15, −3, − , −
OR 𝑆𝑛 =
π‘Ž1 −π‘Ž1
1−π‘Ÿ
π‘Ÿπ‘›
,−
3
sum = −468.744
125
1 7
=
−375−(−375)(5)
1
1−5
= −468.744
GEOMETRIC SERIES (on Reference Sheet)
To find the sum of a certain number of terms of a geometric sequence:
π‘Ž1 − π‘Ž1 π‘Ÿ 𝑛
𝑆𝑛 =
1−π‘Ÿ
Where 𝑆𝑛 = the sum of 𝑛 terms, π‘Ž1 = the first term, and π‘Ÿ = the common ratio
1. Find the sum of the first 8 terms of the sequence: −5, 15, −45, 135, …
ο‚· Why is this a geometric sequence/series? Because you are multiplying
by −3
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
𝑆8 =
−5−(−5)(−3)8
1−(−3)
= 8200
25
2. Given the geometric series: {−6 + 9 −
27
2
+ β‹― } find the sum of the first 11
terms.
ο‚· How do you find the common ratio? −6, 9, −
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
=
−6−(−6)(−1.5)11
1−(−1.5)
=−
27
2
,…
π‘Ÿ=
9
−6
= −1.5
107517
512
3. Find the sum of the following summation notations by using the geometric
series formula.
π‘˜−1
∑10
π‘˜=1 4(3)
𝑆𝑛 =
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1 form so it is geometric!
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
=
4−4(3)10
1−3
= 118096
GEOMETRIC SERIES (need to find number of terms)
ο‚· If you need to find the number of terms in a geometric series, use the
geometric sequence formula to find "𝒏".
GEOMETRIC SEQUENCE:
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
GEOMETRIC SERIES
π‘Ž1 − π‘Ž1 π‘Ÿ 𝑛
𝑆𝑛 =
1−π‘Ÿ
4. Calculate the sum of the geometric series: 4 + 12 + β‹― + 972.
π‘Žπ‘› = π‘Ž1 π‘Ÿ
𝑛−1
972 = 4(3)
𝑛−1
243 = 3𝑛−1
𝑆𝑛 =
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
4−4(3)6
1−3
𝑆𝑛 = 1456
35 = 3𝑛−1
5=𝑛−1
𝑛=6
26
1
1
1
1
2
4
8
512
5. Calculate the sum of the geometric series: 1 − + − + β‹― −
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
−
−
−
𝑆𝑛 =
1 𝑛−1
1
512
1
512
19
= 1 (− )
=−
=−
2
𝑆𝑛 =
2
1𝑛−1
𝑆𝑛 =
2
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
1
1−1(− )10
2
1
1−(−2)
341
512
1𝑛−1
2
9=𝑛−1
𝑛 = 10
6. Calculate the sum of the geometric series: 1 + √3 + 3 + 3√3 + β‹― + 243
𝑆10
1 10
1 − 1 (− )
341
2
=
=
1
512
1 − (− )
2
7*. The sum of a geometric series with four terms is 60, and the common ratio is
1
π‘Ÿ = . Find the first term.
2
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
=
π‘Ž1 (π‘Ž−π‘Ÿ 𝑛 )
𝑆𝑛 = 60, 𝑛 = 4, π‘Ÿ =
1−π‘Ÿ
1
2
14
π‘Ž1 (1 − )
2
60 =
1
1−
2
1
30 = π‘Ž1 (1 − )
16
π‘Ž1 = 32
27
Homework 12.5: Geometric Series
1. Find the sum of the first 8 terms of the sequence: −12, −48, −192, −768, …
𝑖−1
2. Find the sum ∑12
by using the geometric series formula.
𝑖=1 6(−2)
3. Calculate the sum of the geometric series: 0.8 + 0.64 + 0.512 + β‹― + 0.32768
4. A geometric sequence has all its terms positive. The first term is 810 and the
fifth term is 10. Find the common ratio.
5. The sum of a geomtric series with seven terms is −10,922, and the common
ratio is π‘Ÿ = 4, Find the first term.
6. The sixth term of an arithmetic sequence is 24. The common difference is 8.
Calculate the first term of the sequence.
28
Lesson 12.6: Geometric Series Word Problems
Learning Goal: How do we use the geometric series formula when working with
word problems?
Do Now: Answer the following question in order to prepare for today’s lesson.
1. What are the formulas for exponential growth and decay? What do the
variables represent in each formula?
Growth: 𝐴 = 𝑃(1 + π‘Ÿ)𝑑
Decay: 𝐴 = 𝑃(1 − π‘Ÿ)𝑑
𝐴 = final amount, 𝑃 = initial amount, π‘Ÿ = rate, 𝑑 = time
𝐴 = final amount, 𝑃 = initial amount, π‘Ÿ = rate, 𝑑 = time
2. A computer valued at $6500 depreciates at the rate of 14.3% per year.
a. Write a function that models the value of the computer after 𝑛 years.
Decay: 𝐴 = 𝑃(1 − π‘Ÿ)𝑑
Geometric!
𝐴 = 6500(1 − .143)𝑛
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
𝐴 = 6500(.857)𝑛
b. Find the value of the computer after three years. 𝑛 = 3
𝐴 = 6500(.857)3 = $4091.25
1
3. A geometric sequence has a first term of 36 and a common ration of .
3
a. Write a geometric series formula, 𝑆𝑛 , to represent this sum over 𝑛 terms.
𝑆𝑛 =
π‘Ž1 −π‘Ž1
π‘Ÿπ‘›
1−π‘Ÿ
1 𝑛
=
36−36(3)
1
1−3
b. Use this formula to find the sum of the first 15 terms.
1 15
𝑆15 =
36−36(3)
1
1−3
= 53.99999624
29
GEOMETRIC SEQUENCES WITH PERCENTAGES
GEOMETRIC SEQUENCE FORMULA
To find any term of a geometric sequence:
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
Where π‘Žπ‘› = the 𝑛th term, π‘Ž1 = the first term, and π‘Ÿ = the common ratio
ο‚· If a question refers to a percent, this means you are dealing with a
geometric sequence.
ο‚· When given a percent, the common ratio is the percent remaining of the
previous term.
ο‚· Growth: 𝐴 = 𝑃(1 + π‘Ÿ)𝑑 & Decay: 𝐴 = 𝑃(1 − π‘Ÿ)𝑑
Example: Identify the common ratio in each situation.
a. A certain water filtration system can remove 70% of the contaminants each
time a sample of water is passed through it. Decay!
1 − π‘Ÿ = 1 − .70 = .30 = π‘Ÿ
b. John’s salary earns an increase of 4% each successive year. Growth!
1 + π‘Ÿ = 1 + .04 = 1.04 = π‘Ÿ
c. A basketball is dropped vertically. The height of each subsequent bounce is
90% of the previous bounce. Decay! Decreases by 10%
1 − π‘Ÿ = 1 − .10 = .90 = π‘Ÿ
1. A fan is running at 10 revolutions per second. After it is turned off, its speed
decreased at a rate of 75% per second. Find an explicit formula for the
sequence that represents the number of revolutions after 𝑛 seconds. Geometric
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
π‘Ž1 = 10
π‘Ÿ = 1 − .75 = .25
π‘Žπ‘› = 10(. 25)𝑛−1
30
2. Suppose you drop a tennis ball from a height of 15 feet. After the ball hits the
floor, it rebounds to 85% of its previous height. Write an explicit formula for the
sequence. Geometric Decrease by 15%
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1
π‘Ž1 = 15
π‘Ÿ = 1 − .15 = .85
π‘Žπ‘› = 15(. 85)𝑛−1
GOMETRIC SERIES WORD PROBLEMS
𝑆𝑛 = the sum of 𝑛 terms,
π‘Ž1 = the first term, and
π‘Ÿ = the common ratio
GEOMETRIC SERIES
𝑆𝑛 =
π‘Ž1 − π‘Ž1 π‘Ÿ
1−π‘Ÿ
𝑛
3. George has taken a job with a starting salary of $50,000 and receives an
annual raise of 2%. Write a geometric series formula, 𝑠𝑛 , for George’s total
earnings over 𝑛 years.
𝑆𝑛 =
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
π‘Ž1 = 50000
1−π‘Ÿ
π‘Ÿ = 1 + .02 = 1.02
𝑛 =?
50000−50000(1.02)𝑛
1−1.02
Use this formula to find George’s total earnings for his first 12 years of working,
to the nearest cent.
𝑆𝑛 =
50000−50000(1.02)12
1−1.02
= $670604.49
4. The first swing of a pendulum travels 40 centimeters. Each subsequent swing
travels 95% as far as the previous swing. Write a geometric series formula, 𝑠𝑛 ,
for the pendulum’s total distance over 𝑛 swings. It decreases by 5%
𝑆𝑛 =
𝑆𝑛 =
π‘Ž1 −π‘Ž1 π‘Ÿ 𝑛
1−π‘Ÿ
π‘Ž1 = 40
π‘Ÿ = 1 − .05 = .95
𝑛 =?
40−40(.95)𝑛
1−.95
Use this formula to find the pendulum’s total distance after the 30th swing, to the
nearest tenth of a centimeter.
𝑆𝑛 =
40−40(.95)30
1−.95
= 628.3
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PUTTING IT ALL TOGETHER
5. A car with an original price of $30,000 depreciates by 30% each year.
a. Write an explicit formula for the price of the car after 𝑛 years.
Decreases by 30%
π‘Žπ‘› = π‘Ž1 π‘Ÿ 𝑛−1 = 30000(.70)𝑛−1
b. Write a recursive formula, the price of the car after 𝑛 years.
π‘Ž1 = 30000
π‘Žπ‘› = 0.7π‘Žπ‘›−1
c. Write a geometric series formula, 𝑠𝑛 , for the car’s total price over 𝑛 years.
𝑆𝑛 =
30000−30000(.70)𝑛
1−.70
6. Kristin wants to increase her running endurance. According to experts, a
gradual mileage increase of 10% per week can reduce the risk of injury. If Kristin
runs 8 miles in week one, which expression can help her find the total number of
miles she will have run over the course of her 6-week training program?
(1) ∑6𝑛=1 8(1.10)𝑛−1
(2) ∑6𝑛=1 8(1.10)𝑛
(3)
8−8(1.10)6
0.90
(4)
8−8(0.10)𝑛
1.10
1 + .10 = 1.10
32
Homework 12.6: Geometric Series Word Problems
1. Alexa earns $33,000 in her first year of teaching and earns a 4% increase in
each successive year. Write a geometric series formula, 𝑆𝑛 , for Alexa’s total
earnings over 𝑛 years. Use this formula to find Alexa’s total earnings for her first
15 years of teaching, to the π‘›π‘’π‘Žπ‘Ÿπ‘’π‘ π‘‘ 𝑐𝑒𝑛𝑑.
2. The tuition fees for the first three years of high school are given in the table
below. These tuition fees form a geometric sequence.
(a) Find the common ratio, π‘Ÿ, for this sequence.
(b) If fees continue to rise at the same rate, calculate (to the nearest dollar) the
total cost of tuition fees for the first six years of high school.
33
3. A basketball is dropped vertically. It reaches a height of 2 cm on the first
bounce. The height of each subsequent bounce is 90% of the previous bounce.
(a) What height does it reach on the 8th bounce?
(b) What is the total vertical distance travelled by the ball between the first and
sixth time the ball hits the ground?
4. The sum of a geometric series with seven terms is 13,116, and the common
ratio is π‘Ÿ = 3. Find the first term.
5. A theater has 35 seats in the first row. Each row has four more seats than the
row before it. Which expression represents the number of seats in the 𝑛th row?
(1) 35 + (𝑛 + 4)
(2) 35 + (4𝑛)
(3) 35 + (𝑛 + 1)(4)
(4) 35 + (𝑛 − 1)(4)
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