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9.1-9.3
Sequences and Series
Quick Review
Evaluate each expression when a  3, r  2, n  4 and d  2.
1. a  ( n  1) d
2. a  r
Find a .
n 1
10
k 1
k
4. a  2  3
3. a 
k
k 1
k
5. a  a  3 and a  10
k
k 1
9
Quick Review Solutions
Evaluate each expression when a  3, r  2, n  4 and d  2.
1. a  ( n  1) d 9
2. a  r
24
Find a .
n 1
10
k 1
k
4. a  2  3
3. a 
k
k 1
k
11
10
39,366
5. a  a  3 and a  10
k
k 1
9
13
What you’ll learn about
•
•
•
•
Infinite Sequences
Limits of Infinite Sequences
Arithmetic and Geometric Sequences
Sequences and Graphing Calculators
… and why
Infinite sequences, especially those with
finite limits, are involved in some key
concepts of calculus.
Limit of a Sequence
Let a  be a sequence of real numbers, and consider lim a .
n
n 
n
If the limit is a finite number L, the sequence converges and L
is the limit of the sequence. If the limit is infinite or nonexistent,
the sequence diverges.
Example Finding Limits
of Sequences
Determine whether the sequence converges or diverges. If it converges,
give the limit.
2 1 2
2
2,1, , , ,..., ,...
3 2 5
n
Example Finding Limits of
Sequences
Determine whether the sequence converges or diverges. If it converges,
give the limit.
2 1 2
2
2,1, , , ,..., ,...
3 2 5
n
2
lim  0, so the sequence converges to a limit
n
n 
Arithmetic Sequence
A sequence a  is an arithmetic sequence if it can be written in the
n
form a, a  d , a  2d ,..., a  ( n  1) d ,... for some constant d .
The number d is called the common difference.
Each term in an arithmetic sequence can be obtained recursively from
its preceding term by adding d : a  a  d (for all n  2).
n
n 1
Example Arithmetic
Sequences
Find (a) the common difference, (b) the tenth term, (c) a
recursive rule for the
nth term, and (d) an explicit rule for the nth term.
-2, 1, 4, 7, …
Example Arithmetic
Sequences
Find (a) the common difference, (b) the tenth term, (c) a
recursive rule for the
nth term, and (d) an explicit rule for the nth term.
-2, 1, 4, 7, …
(a) The common difference is 3.
(b) a  2  (10  1)3  25
10
(c) a  2 a  a  3 for all n  2
1
n
n 1
(d) a  2  3(n  1)  3n  5
n
Geometric Sequence
A sequence a
n

is a geometric sequence if it can be written in the
form a, a  r , a  r ,..., a  r ,... for some nonzero constant r .
2
n 1
The number r is called the common ratio.
Each term in a geometric sequence can be obtained recursively from
its preceding term by multiplying by r : a  a  r (for all n  2).
n
n 1
Example Defining
Geometric Sequences
Find (a) the common ratio, (b) the tenth term, (c) a recursive
rule for the
nth term, and (d) an explicit rule for the nth term.
2, 6, 18,…
Example Defining
Geometric Sequences
Find (a) the common ratio, (b) the tenth term, (c) a recursive
rule for the
nth term, and (d) an explicit rule for the nth term.
2, 6, 18,…
(a) The ratio is 3.
(b) a  2  3  39,366
10 1
10
(c) a  2 and a  3a
1
n
(d) a  2  3 .
n 1
n
n 1
for n  2.
The Fibonacci Sequence
The Fibonacci sequences can be defined recursively by
a 1
1
a 1
2
a  a a
n
n2
n 1
for all positive integers n  3.
9.1-9.3
Sequences and Series (cont.)
Quick
Review
a  is an arithmetic sequence. Use the given information to find a .
n
10
1. a  5; d  4
1
2. a  5; d  2
3
a  is a geometric sequence. Use the given information to find a
n
10
3. a  5; r  4
1
4. a  5; r  4
3
5. Find the sum of the first 3 terms of the sequence n  .
2
.
Quick Review Solutions
a 
n
is an arithmetic sequence. Use the given information to find a .
10
1. a  5; d  4
41
2. a  5; d  2
19
1
3
a 
n
is a geometric sequence. Use the given information to find a .
10
3. a  5; r  4 1,310,720
1
4. a  5; r  4
3
81,920
5. Find the sum of the first 3 terms of the sequence n  . 14
2
What you’ll learn about
•
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•
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Summation Notation
Sums of Arithmetic and Geometric Sequences
Infinite Series
Convergences of Geometric Series
… and why
Infinite series are at the heart of integral
calculus.
Summation Notation
In summation notation, the sum of the terms of the sequence a , a ,..., a 
1
is denoted  a which is read "the sum of a from k  1 to n."
n
k 1
k
k
The variable k is called the index of summation.
2
n
Sum of a Finite
Arithmetic Sequence
Let a , a ,..., a  be a finite arithmetic sequence with common difference d .
1
2
n
Then the sum of the terms of the sequence is
 a  a  a  ...  a
n
k 1
k
1
2
n
a a 
 n

 2 
n
  2a  (n  1)d 
2
1
n
1
Example Summing the Terms of
an Arithmetic Sequence
A corner section of a stadium has 6 seats along the front row. Each
successive row has 3 more seats than the row preceding it. If the top
row has 24 seats, how many seats are in the entire section?
Example Summing the Terms
of an Arithmetic Sequence
A corner section of a stadium has 6 seats along the front row. Each
successive row has 3 more seats than the row preceding it. If the top
row has 24 seats, how many seats are in the entire section?
The number of seats in the rows form an arithmetic sequence with
a  6, a  24, and d  3. Solving
1
n
a  a  (n  1)d
n
1
24  6  3(n  1)
n7
Apply the Sum of a Finite Sequence Theorem:
 6  24 
Sum of chairs  7 
  105. There are 105 seats in the section.
 2 
Sum of a Finite
Geometric Sequence
Let a , a ,..., a  be a finite geometric sequence with common ratio r.
1
2
n
Then the sum of the terms of the sequence is
 a  a  a  ...  a
n
k 1
k
1

2
a 1  r
1
1 r
n

n
Infinite Series
An infinite series is an expression of the form

 a  a  a  ...  a  ...
k 1
k
1
2
n
Sum of an Infinite
Geometric Series

The geometric series  a  r
k 1
k 1
converges if and only if | r | 1.
a
If it does converge, the sum is
.
1 r
Example Summing Infinite
Geometric Series
Determine whether the series converges. If it converges, give the sum.
 2  0.25 

k 1
k 1
Example Summing Infinite
Geometric Series
Determine whether the series converges. If it converges, give the sum.
 2  0.25 

k 1
k 1
Since |r | 0.25  1, the series converges.
a
2
8
The sum is

 .
1  r 1  0.25 3
9.4
Mathematical Induction
Quick Review
1. Expand the product k ( k  2)( k  4).
Factor the polynomial.
2. n  7 n  10
3. n  3n  3n  1
2
3
2
x
.
x 1
5. Find f (t  1) given f ( x)  x  1.
4. Find f (t ) given f ( x) 
2
Quick Review Solutions
1. Expand the product k (k  2)(k  4). k  6k  8k
3
2
Factor the polynomial.
 n  2  n  5 
 3n  1  n  1
2. n  7 n  10
2
3. n  3n
3
2
3
x
t
.
x 1 t 1
5. Find f (t  1) given f ( x )  x  1. t  2t  2
4. Find f (t ) given f ( x ) 
2
2
What you’ll learn about
• The Tower of Hanoi Problem
• Principle of Mathematical Induction
• Induction and Deduction
… and why
The principle of mathematical induction is a
valuable technique for proving combinatorial
formulas.
The Tower of Hanoi
Solution
The minimum number of moves required
to move a stack of n washers in a Tower
of Hanoi game is 2n – 1.
Principle of Mathematical
Induction
Let Pn be a statement about the integer
n. Then Pn is true for all positive
integers n provided the following
conditions are satisfied:
1. (the anchor) P1 is true;
2. (inductive step) if Pk is true, then Pk+1
is true.
9.5
The Binomial Theorem
Quick Review
Use the distributive property to expand the binomial.
1.  x  y 
2
2. (a  2b)
3. (2c  3d )
4. (2 x  y )
2
2
5.  x  y 
3
2
Quick Review Solutions
Use the distributive property to expand the binomial.
1.  x  y 
x  2 xy  y
2
2
2
2. ( a  2b) a  4ab  4b
3. (2c  3d ) 4c  12cd  9d
4. (2 x  y ) 4 x  4 xy  y
2
2
2
2
2
2
5.  x  y 
3
2
2
2
x  3 x y  3 xy  y
3
2
2
3
What you’ll learn about
•
•
•
•
Powers of Binomials
Pascal’s Triangle
The Binomial Theorem
Factorial Identities
… and why
The Binomial Theorem is a marvelous study in
combinatorial patterns.
Binomial Coefficient
The binomial coefficients that appear in the expansion of (a  b)
n
are the values of C for r  0,1, 2,3,..., n.
n
r
A classical notation for C , especially in the context of binomial
n
r
n
coefficients, is   . Both notations are read "n choose r."
r
Example Using nCr to
Expand a Binomial
Expand  a  b  , using a calculator to compute the binomial coefficients.
4
Example Using nCr to
Expand a Binomial
Expand  a  b  , using a calculator to compute the binomial coefficients.
4
0,1, 2,3, 4 into the calculator to find the binomial
coefficients for n  4. The calculator returns the list 1,4,6,4,1.
Enter 4 C
n
r
Using these coefficients, construct the expansion:
 a  b   a  4a b  6a b  4ab  b .
4
4
3
2
2
3
4
The Binomial Theorem
For any positive integer n,
n
n
n
 n
 a  b     a    a b  ...    a b  ...    b ,
0
1
r
 n
n
n!
where    C 
.
r !(n  r )!
r
n
n
n
r
n 1
nr
r
n
Basic Factorial Identities
For any integer n  1, n !  n  n  1!
For any integer n  0,  n  1!   n  1 n !
9.6
Counting Principles
Quick Review
Give the number of objects described.
1. The number of cards in a standard deck.
2. The number of face cards in a standard deck.
3. The number of vertices of a octogon.
4. The number of faces on a cubical die.
5. The number of possible totals when two dice are rolled.
Quick Review Solutions
Give the number of objects described.
1. The number of cards in a standard deck. 52
2. The number of face cards in a standard deck. 12
3. The number of vertices of a octogon. 8
4. The number of faces on a cubical die. 6
5. The number of possible totals when two dice are rolled. 11
What you’ll learn about
•
•
•
•
•
•
Discrete Versus Continuous
The Importance of Counting
The Multiplication Principle of Counting
Permutations
Combinations
Subsets of an n-Set
… and why
Counting large sets is easy if you know the correct
formula.
Multiplication Principle of
Counting
If a procedure P has a sequence of stages S , S ,..., S and if
1
2
n
S can occur in r ways,
1
1
S can occur in r ways
2
2
S can occur in r ways,
n
n
then the number of ways that the procedure P can occur is the
product rr ...r .
1
2
n
Example Using the
Multiplication Principle
If a license plate has four letters followed by three
numerical digits. Find the number of different
license plates that could be formed if there is no
restriction on the letters or digits that can be used.
Example Using the
Multiplication Principle
If a license plate has four letters followed by three
numerical digits. Find the number of different
license plates that could be formed if there is no
restriction on the letters or digits that can be used.
You can fill in the first blank 26 ways, the second blank 26
ways, the third blank 26 ways, the fourth blank 26 ways, the
fifth blank 10 ways, the sixth blank 10 ways, and the seventh
blank 10 ways. By the Multiplication Principle, there are
26×26×26×26×10×10×10 = 456,976,000 possible license
plates.
Permutations of an n-Set
There are n! permutations of an n-set.
Example Distinguishable
Permutations
Count the number of different 8-letter “words” that
can be formed using the letters in the word
COMPUTER.
Example Distinguishable
Permutations
Count the number of different 8-letter “words” that
can be formed using the letters in the word
COMPUTER.
Each permutation of the 8 letters forms a different word.
There are 8! = 40,320 such permutations.
Distinguishable
Permutations
There are n ! distinguishable permutations of an n-set containing n
distinguishable objects.
If an n-set contains n objects of a first kind, n objects of a second
1
2
kind, and so on, with n  n  ...  n  n, then the number of
1
2
k
distinguishable permutations of the n-set is
n!
.
n !n !n ! n !
1
2
3
k
Permutations Counting
Formula
The number of permutations of n objects taken r at a time is
n!
denoted P and is given by P 
for 0  r  n.
 n  r !
n
r
n
If r  n, then P  0.
n
r
r
Combination Counting
Formula
The number of combinations of n objects taken r at a time is
n!
denoted C and is given by C 
for 0  r  n.
r ! n  r  !
n
r
n
If r  n, then C  0.
n
r
r
Example Counting
Combinations
How many 10 person committees can be formed
from a group of 20 people?
Example Counting
Combinations
How many 10 person committees can be formed
from a group of 20 people?
Notice that order is not important. Using combinations,
20!
C 
 184,756.
10! 20  10 !
20
10
There are 184,756 possible committees.
Formula for Counting
Subsets of an n-Set
n
There are 2 subsets of a set with n objects (including the
empty set and the entire set).
9.7
Probability
Quick Review
How many outcomes are possible for the following experiments.
1. Two coins are tossed.
2. Two different 6-sided dice are rolled.
3. Two chips are drawn simultaneously without replacement from
a jar with 8 chips.
4. Two different cards are drawn from a standard deck of 52.
C
5. Evaluate without using a calculator.
C
4
2
8
2
Quick Review Solutions
How many outcomes are possible for the following experiments.
1. Two coins are tossed. 4
2. Two different 6-sided dice are rolled. 36
3. Two chips are drawn simultaneously without replacement from
a jar with 8 chips. 28
4. Two different cards are drawn from a standard deck of 52. 1326
C
5. Evaluate without using a calculator.
3/14
C
4
2
8
2
What you’ll learn about
•
•
•
•
•
Sample Spaces and Probability Functions
Determining Probabilities
Venn Diagrams and Tree Diagrams
Conditional Probability
Binomial Distributions
… and why
Everyone should know how mathematical the
“laws of chance” really are.
Probability of an Event
(Equally Likely Outcomes)
If E is an event in a finite, nonempty sample space S of equally likely
outcomes, then the probability of the event E is
the number of outcomes in E
P( E ) 
.
the number of outcomes in S
Probability Distribution for
the Sum of Two Fair Dice
Outcome
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
Example Rolling the Dice
Find the probability of rolling a sum divisible by 4 on
a single roll of two fair dice.
Example Rolling the Dice
Find the probability of rolling a sum divisible by 4 on
a single roll of two fair dice.
The event E consists of the outcomes 4,8,12. To get the probability
of E we add up the probabilities of the outcomes in E:
3
5
1
9 1
P( E ) 
 

 .
36 36 36 36 4
Probability Function
A probability function is a function P that assigns a real number
to each outcome in a sample space S subject to the following conditions:
1. 0  P(O)  1;
2. the sum of the probabilities of all outcomes in S is 1;
3. P()  0.
Probability of an Event
(Outcomes not Equally Likely)
Let S be a finite, nonempty sample space in which every outcome
has a probability assigned to it by a probability function P. If E is
any event in S , the probability of the event E is the sum of the
probabilities of all the outcomes contained in E.
Strategy for Determining
Probabilities
1. Determine the sample space of all possible outcomes. When possible,
choose outcomes that are equally likely.
2. If the sample space has equally likely outcomes, the probability of an
the number of outcomes in E
event E is determined by counting: P ( E ) 
.
the number of outcomes in S
3. If the sample space does not have equally likely outcomes, determine
the probability function. (This is not always easy to do.) Check to be sure
that the conditions of a probability function are satisfied. Then the
probability of an event E is determined by adding up the probabilities
of all the outcomes contained in E.
Example Choosing
Chocolates
Dylan opens a box of a dozen chocolate cremes and offers
three of them to Russell. Russell likes vanilla cremes the
best, but all the chocolates look alike on the outside. If five
of the twelve cremes are vanilla, what is the probability
that all of Russell’s picks are vanilla?
Example Choosing
Chocolates
Dylan opens a box of a dozen chocolate cremes and offers
three of them to Russell. Russell likes vanilla cremes the
best, but all the chocolates look alike on the outside. If five
of the twelve cremes are vanilla, what is the probability
that all of Russell’s picks are vanilla?
The experiment in question is the selection of three chocolates,
without regard to order, from a box of 12. There are C  220
12
3
outcomes of this experiment. The event E consists of all possible
combinations of 3 that can be chosen, without regard to order, from
the 5 vanilla cremes available. There are C  10 ways.
5
Therefore, P( E )  10 / 220  1/ 22.
3
Multiplication Principle of
Probability
Suppose an event A has probability p1
and an event B has probability p2 under
the assumption that A occurs. Then the
probability that both A and B occur is
p1p2.
Example Choosing
Chocolates
Dylan opens a box of a dozen chocolate cremes and offers
three of them to Russell. Russell likes vanilla cremes the
best, but all the chocolates look alike on the outside. If five
of the twelve cremes are vanilla, what is the probability
that all of Russell’s picks are vanilla?
Example Choosing
Chocolates
Dylan opens a box of a dozen chocolate cremes and offers
three of them to Russell. Russell likes vanilla cremes the
best, but all the chocolates look alike on the outside. If five
of the twelve cremes are vanilla, what is the probability
that all of Russell’s picks are vanilla?
The probability of picking a vanilla creme on the first draw is 5/12.
Under the assumption that a vanilla creme was selected in the first
draw, the probability of picking a vanilla creme on the second draw is
4/11. Under the assumption that a vanilla creme was selected in the first
and second draw, the probability of picking a vanilla creme on the third
draw is 3/10. By the Multiplication Principle, the probability of picking
5 4 3
60
1
a vanilla creme on all three picks is
  
 .
12 11 10 1320 22
Conditional Probability
Formula
If the event B depends on the event A, then P( B | A) 
P( A and B)
.
P( A)
Binomial Distribution
Suppose an experiment consists of n-independent repetitions of an
experiment with two outcomes, called "success" and "failure." Let
P(success)  p and P(failure)  q. (Note that q  1  p.)
Then the terms in the binomial expansion of ( p  q) give the respective
probabilities of exactly n, n  1,..., 2, 1, 0 successes.
Number of successes out of
Probability
n independent repetitions
n
p
n
n
n 1
1
0
 n 
 n  1 p q


n 1
n
 r  pq
 
q
n
n 1
Example Shooting Free
Throws
Suppose Tommy makes 92% of his free throws. If he shoots
15 free throws, and if his chance of making each one is
independent of the other shots, what is the probability that he
makes all 15?
Example Shooting Free
Throws
Suppose Tommy makes 92% of his free throws. If he shoots
15 free throws, and if his chance of making each one is
independent of the other shots, what is the probability that he
makes all 15?
P(15 successes)   0.92   0.286
15
Example Shooting Free
Throws
Suppose Tommy makes 92% of his free throws. If he shoots
15 free throws, and if his chance of making each one is
independent of the other shots, what is the probability that he
makes exactly 10?
Example Shooting Free
Throws
Suppose Tommy makes 92% of his free throws. If he shoots
15 free throws, and if his chance of making each one is
independent of the other shots, what is the probability that he
makes exactly 10?
 15 
P(10 successes)=    0.92   0.08   0.00427
10 
10
5