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Chapter 7
College Algebra
7.1 Sequences and Series
 An infinite sequence is a function having for its
domain the set of positive integers
 {1,2,3,4,5,…}
 A finite sequence is a function having for its domain a
set of the positive integers,
 {1,2,3,4,5,…,n},
 for some positive integer n.
Example for a Sequence
 Find the first four terms and the 23rd term of the
sequence whose general term is given by
an = (-1)nn2
 a1 = (-1)1(1)2 = -1
a2 = (-1)2(2)2 = 4
a3 = (-1)3(3)2 = -9
a4 = (-1)4(4)2 = 16
a23 = (-1)23(23)2 = -529
 Predict the general term of the sequence {2,4,8,…}
 There is a pattern of multiplying by 2 so the general
term for the sequence is 2n
Series
 Given the infinite sequence
a1, a2, a3, …. , an, ….
The sum of the terms
a1+a2+a3+….+an+…
is called an infinite series. A partial sum is the sum of
the first n terms
a1+a2+a3+…+an.
 A partial sum is also called a finite series or nth partial
sum and is denoted Sn
Example
 For the sequence -2, 4 -6, 8, -10, 12, -14,…, find the
following
S1, S4, and S5
 S1 = -2
S4 = -2+4+(-6)+8 = 4
S5 = S4 + (-10) = -6
Sigma Notation Σ
 The Greek letterΣ(sigma) can be used to simplify
notation when the general term of a sequence is a
formula.
 For example when a sequence can be given by the
formula 2k + 1 {3, 5, 7, …}, the sum of the first four
terms can be written as
4
å 2k +1
k=1
Recursive Definitions
 A sequence may be defined recursively or by using a
recursive formula.
 Find the first five terms of the sequence given by
a1 = 5
 a1 = 5
and an+1 = 2an – 3
a2 = 2(5) – 3 = 7
a3 = 2(7) – 3 = 11
for n > 1
7.2 Arithmetic Sequences
and Series
 An sequence is arithmetic if there exists a number d,
called the common difference, such that
an+1 = an + d for any integer n > 1
 The nth term of an arithmetic sequence is given by
an = a1 + (n-1)d
for any integer n > 1
 For the sequence {4, 9, 14, 19, 24, …} find the first
term, the common difference, and the nth term
a1 = 4,
d= 9-4 = 5,
so an = 4 + (n-1)5 = 4 + 5n – 5 = 5n - 1 so an = 5n - 1
Arithmetic Sums
 The sum of the first n terms of an arithmetic sequence
is given by Sn = (n/2) (a1 + an)
 Example: Find the sum of the first 100 natural
numbers:
The sum is 1 + 2 + 3 + 4 + … + 99 + 100
Therefore
S100 = (100/2) ( 1 + 100) = 50 (101) = 5050
Homework
7.3 Geometric Series and
Sequences
 A sequence is geometric if there is a number r, called the
common ratio, such that
an+1 = anr,
for any integer n > 1.
 The nth term of a geometric sequence is given by
an = a1rn-1
for any integer n > 1.
 Find the 7th term of the geometric sequence 4, 20, 100,…
a1 = 4 and r = 5, we are looking for a7
 a7 = 4 (5)7-1 = 4 (5)6 = 62, 500
Geometric Sum Formulas

The sum of the first n terms of a geometric sequence
is given by
Sn = (a1(1 - r n)) / (1 - r)
for any r not equal to 1
 Find the sum of the first 7 terms of the sequence 3, 15,
75, 375, …
a1 = 3 and r = 5
n=7
S7 = (3(1-57)) / (1-5) = 58,593
 The Limit or Sum of an Infinite Geometric Series
When |r|< 1, the limit or sum of an infinite
geometric series is given by
a1
S¥ =
1- r
7.4 Mathematical
Induction
 Mathematical induction is a method of proof which
can be used to try to prove that all statements in an
infinite sequence of statements are true.
 Statements usually have the form: “For all natural
numbers n, Sn”
 where Sn is some mathematical sentence.
 We can prove an infinite sequence of statements Sn by
showing the following
1) Basis step, S1, is true
2) Induction step.For all natural numbers, k, Sk  Sk+1
Example
 Prove: For every natural number n,
1 + 3 + 5 + … + (2n-1) = n2
 First it can be helpful to find Sn, S1, Sk, and Sk+1
Sn: 1 + 3 + 5 + …. + (2n -1) = n2
S1: 1 = 12
Sk: 1 + 3 + 5 + … + (2k -1) = k2
Sk+1: 1 + 3 + 5 + …+ (2k-1) + (2(k+1) – 1) = (k+1)2
 Basis Step: Prove S1: 1 = 12 = 1.
Induction Step:
Sk+1 = 1 + 3 + … + (2k-1) + (2k + 2 – 1) = (k+1)2
Sk+1 =
k2 + 2k + 1
= (k+1)2
Sk+1 =
(k + 1)2
= (k + 1)2
Homework
7.5 Combinatorics:
Permutations
 In order to study probability it is first necessary to
learn about combinatorics, the theory of counting.
 We will consider part of the branch of combinatorics
called permutations
 The study involves order and arragnements
 Example: How many three letter code symbols can be
formed with the letters A, B, and C without repitition
(that is, using each letter only once)?
 ABC, ACB, BAC, BCA, CAB, and CBA
You could find these using a tree diagram
Fundamental Counting
Principle
 Given a combined action, or event, in which the first
action can be performed n1 ways, the second action
can be performed in n2 ways, and so on, the total
number of ways in which the combined action can be
performed is the product
n1n2n3 … nk
 Example: How many 3-letter code symbols can be
formed with the letters A, B, C, D, and E with
repetition (that is allowing the letters to be repeated)?
5(5)(5) = 53 = 125
Permutations
 A permutation of a set of n objects is an ordered
arrangement of all n objects.
 The total number of permutations of n objects,
denoted nPn is given by
nPn = n(n-1)(n-2) … 3(2)(1)
 In how many different ways can 9 packages be placed
in 9 mailboxes, one package per box?
9P9 = 9(8)(7)(6)(5)(4)(3)(2)(1) = 362, 880
Factorial Notation
 The answer to the last problem 9(8)(7)(6)(5)(4)(3)(2)1
can also be written as 9! read as “9 factorial”.
 For any natural number n,
n! = n ( n - 1) ( n - 2) … 3 (2) (1)
 For the number 0, 0! = 1
 Therefore nPn = n!
 For any natural number n, n! = n (n-1)!
 A permutation of a set of n objects taken k at a time is
an ordered arrangement of k objects taken from the set
Number of Permutations
Taken
k
at
a
Time
The number of permutations of a set of n objects taken k at

a time denoted nPk is given by
= n ( n – 1) ( n – 2) …. (n – (k – 1))
= n! / (n – k)!
 The flags of many nations consist of three vertical stripes.
For example the flag of Ireland has green, white, and
orange vertical stripes. How many different flags of 3 colors
can be made without repetition of colors of in the flag?

9! / 6! = 9 (8) (7) = 504
Repetitions
 In how many distinguishable ways can the letters of
the word CINCINNATI be arranged?
 There are 2C’s, 3 I’s, 3N’s, 1 A, and 1 T for a total of
ten letters:
 10! / (2! 3! 1! 1!) = 50,400
7.6 Combinatorics:
Combinations
 Find all the combinations of 3 letters taken from the
set of 5 letters {A, B, C, D, E}:
{ABC} {ABE} {ABD} {ACD} {ACE} {ADE}
{BCD} {BCE} {BDE} {CDE}
 There are 10 combinations of the 5 letters taken 3 at a
time.
 Notice that {ABC} is consider the same combination
of letters as {ACB}
Vocabulary
 Set A is a subset of set B, denoted A Í B, if every
element of A is an element of B.
 A combination containing k objects is a subset
containing k objects
 The number of combinations of n objects taken k at a
time is denoted as nCk = (nPk / k!) = (n! / (k!(n-k)!) )
 Binomial Coefficient Notation nCk = æ n ö
ç
÷
è k ø
Subsets
 Subsets of size k and of size n – k
ænö
æn
ö
ç ÷=ç
÷
èk ø
èn - k ø
The number of subsets of size k of a set with n objects
is the same as the number of subsets of size n – k. The
number of combinations of n objects taken k at a time
is the same as the number of combinations of n
objects taken n – k at a time.
Example
 Run by the state of Michigan, Classic Lotto 47 is a
twice-weekly lottery game with jackpots starting at $1
million. For a wager of $1, a player can choose 6
numbers from 1 through 47. If the numbers match
those drawn by the state, the player wins.
 A) How many 6-number combinations are there
 B) Suppose it takes you 10 min to pick your numbers
and buy a game ticket. How many tickets can you buy
in 4 days?
 C) How many people would you have to hure for 4
days to buy tickets with all the possible combinations
and ensure that you win?
Example
 A) How many 6-number combinations are there?

No order is implied so (47! / 6! 41!) = 10,737,573
 B) Suppose it takes you 10 min to pick your numbers and
buy a game ticket. How many tickets can you buy in 4
days?

First we find the number of minutes in four days
5760 minutes
Thus you could buy 5760/10 or 576 tickets in 4 days
 C) How many people would you have to hire for 4 days to
buy tickets with all the possible combinations and ensure
that you win?

You would need to hire (10,737,573 / 576) or 18,642 people
to buy tickets with all combinations to win.
Homework
7.7 Binomial Theorem
 For any binomial a + b and any natural number n,
(a+b)n = c0anb0 + c1an-1b + c2an-2b2 + …. + cna0bn
where the numbers ck are from the (n+1)st row of
Pascal’s Triangle
Binomial Theorem
 For any binomial a + b and any natural number n,
(a+b)n =
æ n ö n æ n ö n-1
æ n ö n-2 2
æn
ö n-1 æ n ö n
a
+
a
b
+
a
b
+...
+
ç ÷
ç ÷
ç ÷
ç
÷ ab + ç ÷ b
0
1
2
n
-1
è ø
è ø
è ø
è
ø
ènø
æ n ö n-k k
= åç ÷ a b
k
k=0 è ø
n
Example
 Expand: (2x + 3y)4 using both versions of the
binomial theorem.
Application
 Finding the (k+1)st term of
(a+b)n
is
æ n ö n-k k
ç ÷a b
èk ø
 Example: Find the 5th term in the expansion of
(2x – 5y)6
 k = 4, a = 2x, b = -5y, n = 6
æ6ö
2
4
2 4
ç ÷ (2x) (-5y) = 37500x y
è 4ø
Subsets
 The total number of subsets of a set with n elements is
2n.
 The set of {A,B,C,D,E} has how many subsets?
The set has 5 elements so 25 or 32 subsets.
 The fast food chain Wendy’s offers the following on a
hamburger ketchup, mustard, mayo, tomato, lettuce,
onion, pickle. How many different ways can they serve
a hamburger?
 27 = 128
7.8 Probability
 There are two different kinds of probability, experimental
and theoretical.

Experimental probability is discovered by the observation
and study of data and is quite common and very useful.

Theoretical probability is determined by reasoning
mathematically.
 Probability Properties

A) If an event E cannot occur, then P(E) = 0

B) If an event E is certain to occur then P(E) = 1

C) The probability that an event E will occur is a number
from 0 to 1: 0 < P(E) < 1
Principle
 Principle P (Experimental)
Given an experiment in which n observations are
made, if a situation, or event, E occurs m times out of
n observations, then we say that the experimental
probability of the event P(E) = m/n
 Principle P (Theoretical)
If an event E can occur m ways out of n possible
equally likely outcomes of a sample space S, then the
theoretical probability of the event P(E) = m/n
Examples
 Suppose a die is rolled. Find the following
a) The outcomes
The outcomes are 1, 2, 3, 4, 5, 6.
b) The sample space
The sample space is {1, 2, 3, 4, 5, 6}
c) The probability of rolling a 3
P(3) = 1/6
d) The probability of rolling an even
P(even) = 3/6 = ½
Homework