8.1 Sequences

Report
Warm-Up
• Find the first four
terms of the sequence
given by an  3n  2
Find the first four terms of the
n
sequence given by an  3  (1)
a1  31  2  1
a1  3  (1)1  3 1  2
a2  3 2  2  4
a2  3  (1)2  3  1  4
a3  33  2  7
a3  3  (1)3  3 1  2
a4  3 4  2  10
a4  3  (1)4  3  1  4
Copyright © 2007 Pearson Education, Inc.
Slide 8-2
Chapter 8:
Sequences and Series
2015
Copyright © 2007 Pearson Education, Inc.
Slide 8-3
Chapter 8: Sequences, Series, and Probability
8.1 Sequences and Series
8.2 Arithmetic Sequences and Series
8.3 Geometric Sequences and Series
8.4 Mathematical Induction
8.5 The Binomial Theorem
Copyright © 2007 Pearson Education, Inc.
Slide 8-4
8.1
Sequences
A sequence is a function that has a set of natural
numbers as its domain.
•
•
•
f (x) notation is not used for sequences.
Write an  f (n)
Sequences are written as ordered lists
a1 , a2 , a3 , ...
•
a1 is the first element, a2 the second element,
and so on
Copyright © 2007 Pearson Education, Inc.
Slide 8-6
8.1
Sequences
A sequence is often specified by giving a formula for
the general term or nth term, an.
Example Find the first four terms for the sequence
n 1
an 
n2
Solution
a1  2 / 3,
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a2  3 / 4,
a3  4 / 5,
a4  5 / 6
Slide 8-7
8.1
Graphing Sequences
The graph of a sequence, an, is the graph of the
discrete points (n, an) for n = 1, 2, 3, …
Example Graph the sequence an = 2n.
Solution
Copyright © 2007 Pearson Education, Inc.
Slide 8-8
8.1
Sequences
• A finite sequence has domain the finite set
{1, 2, 3, …, n} for some natural number n.
Example 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
• An infinite sequence has domain
{1, 2, 3, …}, the set of all natural numbers.
Example 1, 2, 4, 8, 16, 32, …
Copyright © 2007 Pearson Education, Inc.
Slide 8-9
8.1
Convergent and Divergent Sequences
• A convergent sequence is one whose terms get
closer and closer to a some real number. The
sequence is said to converge to that number.
• A sequence that is not convergent is said to be
divergent.
Copyright © 2007 Pearson Education, Inc.
Slide 8-10
8.1
Convergent and Divergent Sequences
Example :
1
Find the first 5 terms of the sequence an  .
n
Is the sequence convergent or divergent?
Copyright © 2007 Pearson Education, Inc.
Slide 8-11
8.1
Convergent and Divergent Sequences
1
Solution: The sequence an  converges to 0.
n
The terms of the sequence 1, 0.5, 0.33.., 0.25, …
grow smaller and smaller approaching 0. This can be
seen graphically.
Copyright © 2007 Pearson Education, Inc.
Slide 8-12
8.1
Convergent and Divergent Sequences
Example :
Find the first 6 terms of the sequence
an  n .
2
Is the sequence convergent or divergent?
Copyright © 2007 Pearson Education, Inc.
Slide 8-13
8.1
Convergent and Divergent Sequences
Solution: The sequence an  n2 is divergent.
The terms grow large without bound
1, 4, 9, 16, 25, 36, 49, 64, …
and do not approach any one number.
Copyright © 2007 Pearson Education, Inc.
Slide 8-14
8.1
Convergent and Divergent Sequences
2n  n  3
an  2
3n  2n  5
2
Example Is the sequence
convergent or divergent?
Solution: The sequence converges to 2/3
Copyright © 2007 Pearson Education, Inc.
Slide 8-15
Finding Terms of a Sequence
• Write out the first five terms of the sequence given by
(1) n
an 
2n  1
Solution:
(1)1
1
a1 

 1
2 1  1 2  1
(1)2
1
1
a2 


2  2 1 4 1 3
Copyright © 2007 Pearson Education, Inc.
(1)3
1 1
a3 


2  3 1 6  1 5
(1)4
1
1
a4 


2  4 1 8 1 7
(1)5
1
1
a5 


2  5  1 10  1 9
Slide 8-17
Finding the nth term of a Sequence
• Write an expression for the apparent nth term (an) of each
sequence.
• a. 1, 3, 5, 7, …
b. 2, 5, 10, 17, …
Solution:
a.
n: 1 2 3 4 . . .
n
terms: 1 3 5 7 . . .
an
Apparent pattern: Each term
is 1 less than twice n, which
implies that an  2n  1
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b.
n: 1 2 3 4 … n
terms: 2 5 10 17 … an
Apparent pattern: Each
term is 1 more than the
square of n, which implies
that an  n2  1
Slide 8-18
Additional Example
• Write an expression for the apparent nth term of the sequence:
2 3 4 5
, , , ,...
1 2 3 4
Solution:
n : 1 2 3 4 ... n
2 3 4 5
terms:
... an
1 2 3 4
Apparent pattern: Each term has a
n 1
numerator that is 1 greater than its an 
n
denominator, which implies that
Copyright © 2007 Pearson Education, Inc.
Slide 8-19
Factorial Notation
• If n is a positive integer, n factorial is defined by
n!  1 2  3  4...  (n  1)  n
As a special case, zero factorial is defined as 0! = 1.
Here are some values of n! for the first several nonnegative integers.
Notice that 0! is 1 by definition.
0!  1
3!  1 2  3  6
1!  1
2!  1 2  2
4!  1  2  3  4  24
5!  1 2  3  4  5  120
The value of n does not have to be very large before the value
of n! becomes huge. For instance, 10! = 3,628,800.
Copyright © 2007 Pearson Education, Inc.
Slide 8-20
Finding the Terms of a Sequence Involving
Factorials
• List the first five terms of the sequence given by
2n
an 
n!
Begin with n = 0.
22 4
a2 
 2
2! 2
20 1
a0 
 1
0! 1
23 8 4
a3 
 
3! 6 3
21 2
a1    2
1! 1
24 16 2
a4 


4! 24 3
Copyright © 2007 Pearson Education, Inc.
Slide 8-21
Evaluating Factorial Expressions
• Evaluate each factorial expression. Make sure you use parentheses when
necessary.
a.
b. 2! 6!
c.
n!
8!
2! 6!
3! 5!
(n  1)!
a.
8!
1 2  3  4  5  6  7  8 7  8


 28
2! 6! 1  2 1  2  3  4  5  6
2
b.
2! 6! 1  2 1  2  3  4  5  6 6

 2
3! 5! 1  2  3 1 2  3  4  5 3
c.
n!
1 2  3...(n  1)  n

n
(n  1)! 1 2  3...(n  1)
Copyright © 2007 Pearson Education, Inc.
Slide 8-22
Additional Example
• Write an expression for the apparent nth term of the sequence:
2 2 23 2 4 25
1, 2, , , ,
,...
2 6 24 120
Solution:
n: 1 2
3 4 5
6 ... n
2 2 23 2 4 25
terms: 1, 2,
, , ,
... an
2 6 24 120
Apparent pattern: Each term has a
2n 1
numerator that is 1 greater than its an 
 n  1!
denominator, which implies that
Copyright © 2007 Pearson Education, Inc.
Slide 8-23
Have you ever seen this sequence before?
• 1, 1, 2, 3, 5, 8 …
• Can you find the next three terms in the
sequence?
• Hint: 13,
• 21, 34
• Can you explain this pattern?
Copyright © 2007 Pearson Education, Inc.
Slide 8-24
The Fibonacci Sequence
• Some sequences are defined recursively. To define a sequence
recursively, you need to be given one or more of the first few terms. A
well-known example is the Fibonacci Sequence.
• The Fibonacci Sequence is defined as follows:
a0  1,
a1  1,
ak  ak 2  ak 1, where k  2
Write the first six terms of the Fibonacci Sequence:
a32  a31  a1  a2  1  2  3
a0  1
a42  a41  a2  a3  2  3  5
a 1
1
a22  a21  a0  a1  1  1  2
Copyright © 2007 Pearson Education, Inc.
a52  a51  a3  a4  3  5  8
Slide 8-25
Example
• Write the first five terms of the recursively
defined sequence:
a1  5,
ak 1  ak  3
Solution: 5, 8, 11, 14, 17
Copyright © 2007 Pearson Education, Inc.
Slide 8-26
Homework
• Day 1: Pg. 563 1-9odd, 21-23odd, 35-69 odd
• Day 2: 71-81 odd, 91-103 odd
Copyright © 2007 Pearson Education, Inc.
Slide 8-27
HWQ
Write an expression for the apparent
nth term of the sequence.
1
3
7
15
31
1  ,1  ,1  ,1  ,1  ,...
2
4
8
16
32
Copyright © 2007 Pearson Education, Inc.
Slide 8-28
8.1 Day 2
Series
2015
Copyright © 2007 Pearson Education, Inc.
Slide 8-29
Summation Notation
• Definition of Summation Notation
The sum of the first n terms of a sequence is represented by
n
a
i 1
n
 a1  a2  a3  a4  ...  an
Where i is called the index of summation, n is the upper limit of
summation and 1 is the lower limit of summation.
Copyright © 2007 Pearson Education, Inc.
Slide 8-30
8.1
Series and Summation Notation
A finite series is an expression of the form
n
Sn  a1  a2  a3  ...  an   ai
i 1
and an infinite series is an expression of the form

S  a1  a2  a3  ...  an  ...   ai .
i 1
Copyright © 2007 Pearson Education, Inc.
Slide 8-31
Summation Notation for Sums
• Find each sum.
a.
5
 3i
i 1
b.
6
 (1  k
k 3
2
)
c.
8
1

i 0 i !
Solution:
5
a.
 3i  3(1)  3(2)  3(3)  3(4)  3(5)
i 1
 3(1  2  3  4  5) or 3  6  9  12  15
 45
Copyright © 2007 Pearson Education, Inc.
Slide 8-32
Solutions continued
6
b.
 (1  k
2
)  (1  32 )  (1  42 )  (1  52 )  (1  62 )
k 3
 10  17  26  37
 90
c.
8
1 1 1 1 1 1 1 1 1 1
        

0! 1! 2! 3! 4! 5! 6! 7! 8!
i 0 i !
1 1 1
1
1
1
1
 11  




2 6 24 120 720 5040 40320
 2.71828
Notice that this summation is very close to the irrational
number e  2.718281828 . It can be shown that as more terms
of the sequence whose nth term is 1/n! are added, the sum
becomes closer and closer to e.
Copyright © 2007 Pearson Education, Inc.
Slide 8-33
8.1
Series and Summation Notation
Summation Properties
If a1, a2, a3, …, an and b1, b2, b3, …, bn are two
sequences, and c is a constant, then for every positive
integer n,
n
(a)
n
 c  nc
(b)
i 1
i 1
n
(c)
n
i
 c ai
i 1
n
 (a  b )   a   b
i 1
Copyright © 2007 Pearson Education, Inc.
 ca
n
i
i
i 1
i
i 1
i
Slide 8-34
8.1
Series and Summation Notation
Summation Rules
n(n  1)
i  1  2  ...  n 

2
i 1
n
n(n  1)(2n  1)
2
2
2
2
i  1  2  ...  n 

6
i 1
n
2
2
n
(
n

1)
3
3
3
3
i

1

2

...

n


4
i 1
n
These summation rules can be proven by mathematical induction. 
Copyright © 2007 Pearson Education, Inc.
Slide 8-35
8.1
Series and Summation Notation
Example Use the summation properties to
22
40
14
evaluate (a)  5 (b)  2i (c)  (2i 2  3)
i1
i 1
i 1
Solution
40
(a)
 5  40(5)  200
i1
Copyright © 2007 Pearson Education, Inc.
Slide 8-36
8.1
Series and Summation Notation
22
(b)
 2i
14
(c)
i 1
(b)
(c)
2
(2
i
  3)
i 1
22(22  1)
2i  2 i  2
 506

2
i 1
i 1
22
22
14
14
14
14
14
i 1
i 1
i 1
i 1
i 1
2
2
2
(2
i

3)

2
i

3

2
i

    3
14(14  1)(2 14  1)
2
 14(3)  1988
6
Copyright © 2007 Pearson Education, Inc.
Slide 8-37
Homework
• Day 1: Pg. 563 1-9odd, 21-23odd, 35-69 odd
• Day 2: 71-81 odd, 91-103 odd
Copyright © 2007 Pearson Education, Inc.
Slide 8-38

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