Download Chapter 10 - James Bac Dang

Chapter 10
Sequences, Induction, and
Probability
Sequences and Summation Notation
Definition of a Sequence
An infinite sequence {an} is a function
whose domain is the set of positive
integers. The function values, or terms, of
the sequence are represented by
a1, a2, a3, a4,…, an ,….
Sequences whose domains consist only of
the first n positive integers are called
finite sequences.
Text Example
Write the first four terms of the sequence whose nth term,
or general term, is given:
an = 3n + 4.
Solution We need to find the first four terms of the sequence whose general
term is an = 3n + 4. To do so, we replace each occurrence of n in the formula by
1, 2, 3, and 4.
a1, 1st
term
31+4=3+4=7
a2, 2nd
term
3  2 + 4 = 6 + 4 = 10
a3, 3rd
term
3 3 + 4 = 9 + 4 = 13
a4, 4th
term
3  4 + 4 = 12 + 4 = 16

The first four terms are 7, 10, 13, and 16. The sequence defined by an = 3n + 4
can be written as
7, 10, 13, …, 3n + 4, ….
Factorial Notation
If n is a positive integer, the notation
n! is the product of all positive
integers from n down through 1.
n! = n(n-1)(n-2)…(3)(2)(1)
0! , by definition is 1.
Summation Notation
The sum of the first n terms of a
sequence is represented by the
summation notation

n
a = a1 + a2 + a3 + a4 + ... + an
i =1 i
Where i is the index of summation, n
is the upper limit of summation, and
1 is the lower limit of summation.
Example
• Expand and evaluate the sum:
5
 (5  2i)
i =1
Solution:
5
 (5  2i)
i =1
= 3 + 1 + 1 + 3 + 5
= 5
Example
• Express the sum using summation notation:
1 1 1 1
1+ + + +
2 4 8 16
Solution:
1 1 1 1
1+ + + +
2 4 8 16
5
1
=  i 1
i =1 2
Example
• Express the sum using summation notation:
1 1 1
1
1 + + + + ... + n 1
2 4 8
2
Solution:
1 1 1
1
1 + + + + ... + n 1
2 4 8
2
n
1
=  i 1
i =1 2
Arithmetic
Sequences
Arithmetic Sequences
A mathematical model for the average annual salaries of major league baseball
players generates the following data.
Year
1991
1992
1993
1994
1995
1996
1997
1998
Salary
801,000
892,000
983,000
1,074,000
1,165,000
1,256,000
1,347,000
1,438,000
From 1991 to 1992, salaries increased by $892,000 - $801,000 = $91,000.
From 1992 to 1993, salaries increased by $983,000 - $892,000 = $91,000. If
we make these computations for each year, we find that the yearly salary
increase is $91,000. The sequence of annual salaries shows that each term after
the first, 801,000, differs from the preceding term by a constant amount,
namely 91,000. The sequence of annual salaries
801,000, 892,000, 983,000, 1,074,000, 1,165,000, 1,256,000....
is an example of an arithmetic sequence.
Definition of an Arithmetic
Sequence
• An arithmetic sequence is a sequence in
which each term after the first differs from
the preceding term by a constant amount.
The difference between consecutive terms is
called the common difference of the
sequence.
Text Example
The recursion formula an = an  1  24 models the thousands of Air
Force personnel on active duty for each year starting with 1986. In
1986, there were 624 thousand personnel on active duty. Find the
first five terms of the arithmetic sequence in which a1 = 624 and an
= an  1  24.
Solution The recursion formula an = an  1  24 indicates that each term after
the first is obtained by adding 24 to the previous term. Thus, each year there
are 24 thousand fewer personnel on active duty in the Air Force than in the
previous year.
Text Example cont.
The recursion formula an = an  1  24 models the thousands of Air
Force personnel on active duty for each year starting with 1986. In
1986, there were 624 thousand personnel on active duty. Find the
first five terms of the arithmetic sequence in which a1 = 624 and an
= an  1  24.
Solution
a1 = 624
a2 = a1 – 24 = 624 – 24 = 600
a3 = a2 – 24 = 600 – 24 = 576
a4 = a3 – 24 = 576 – 24 = 552
a5 = a4 – 24 = 552 – 24 = 528
This is given.
Use an = an  1  24 with n = 2.
Use an = an  1  24 with n = 3.
Use an = an  1  24 with n = 4.
Use an = an  1  24 with n = 5.
The first five terms are
624, 600, 576, 552, and 528.
Example
• Write the first six terms of the arithmetic sequence where
a1 = 50 and d = 22
Solution:
•a1 = 50
a2 = 72
a3 = 94
a4 = 116
a5 = 138
a6 = 160
General Term of an Arithmetic
Sequence
• The nth term (the general term) of
an arithmetic sequence with first
term a1 and common difference d
is
• an = a1 + (n-1)d
Text Example
Find the eighth term of the arithmetic sequence whose first term is
4 and whose common difference is 7.
Solution To find the eighth term, as, we replace n in the formula with 8, a1
with 4, and d with 7.
an = a1 + (n  1)d
a8 = 4 + (8  1)(7) = 4 + 7(7) = 4 + (49) = 45
The eighth term is 45. We can check this result by writing the first eight terms
of the sequence:
4, 3, 10, 17, 24, 31, 38, 45.
The Sum of the First n Terms of
an Arithmetic Sequence
• The sum, Sn, of the first n terms of an
arithmetic sequence is given by
n
S n = (a1 + an )
2
in which a1 is the first term and an is the nth
term.
Example
• Find the sum of the first 20 terms of the
arithmetic sequence: 6, 9, 12, 15, ...
Solution:
20
S 20 =
(a1 + a20 )
2
= 10(6 + (6 + 19 * 3)
= 10(6 + 63)
= 10(69) = 690
Example
• Find the indicated sum
15
 (2i  4)
i =1
Solution:
15
 (2i  4)
i =1
15
(( 2) + (26))
2
15
= (24)
2
= 15 *12 = 180
=
Geometric
Sequences
Definition of a Geometric
Sequence
• A geometric sequence is a sequence in
which each term after the first is obtained
by multiplying the preceding term by a
fixed nonzero constant. The amount by
which we multiply each time is called the
common ratio of the sequence.
The Common Ratio
The common ratio, r, is found by dividing any term after the first term by
the term that directly precedes it. In the following examples, the
common ratio is found by dividing the second term by the first term, a2
 a1.
Geometric sequence
1, 5 25, 125, 625...
6, -12, 24, -48, 96....
Common ratio
r=51=5
r = -12  6 = -2
Text Example
Write the first six terms of the geometric sequence with first term 6 and common
ratio 1/3 .
Solution The first term is 6. The second term is 6 
term is 2  1/3, or 2/3. The fourth term is 2/3 
on. The first six terms are
6, 2, 2/3, 2/9, 2/27, 2/81.
1/ , or 2. The third
3
1/ , or 2/ , and so
3
9
General Term of a Geometric
Sequence
• The nth term (the general term) of a
geometric sequence with the first term a1
and common ratio r is
• an = a1 r n-1
Text Example
Find the eighth term of the geometric sequence whose first term is
4 and whose common ratio is 2.
Solution To find the eighth term, a8, we replace n in the formula with 8, a1
with 4, and r with 2.
an = a1r n  1
a8 = 4(2)8  1 = 4(2)7 = 4(128) = 512
The eighth term is 512. We can check this result by writing the first eight terms
of the sequence:
4, 8, 16, 32, 64, 128, 256, 512.
The Sum of the First n Terms of a
Geometric Sequence
The sum, Sn, of the first n terms of a
geometric sequence is given by
a1 (1  r )
Sn =
1 r
n
in which a1 is the first term and r is the
common ratio.
Example
• Find the sum of the first 12 terms of the
geometric sequence:
4, -12, 36, -108, ...
Solution:
a1 (1  r n )
Sn =
1 r
a1 (1  r 12 ) 4(1  (3)12 )
S12 =
=
1 r
1  (3)
4(1  531441)
=
= 531440
4
Value of an Annuity: Interest
Compounded n Times per Year
If P is the deposit made at the end of each
compounding period for an annuity at r percent
annual interest compounded n times per year,
the value, A, of the annuity after t years:
r nt
(1 + )  1
n
A= P
r
n
Example
• To save for retirement, you decide to deposit
$2000 into an IRA at the end of each year for the
next 30 years. If the interest rate is 9% per year
compounded annually, find the value of the IRA
after 30 years.
Solution:
P=2000, r =.09, t = 30, n=1
r nt
.09 1*30
(1 + )  1
(1 +
) 1
n
1
A= P
= 2000
r
.09
n
1
Example cont.
• To save for retirement, you decide to deposit
$2000 into an IRA at the end of each year for the
next 30 years. If the interest rate is 9% per year
compounded annually, find the value of the IRA
after 30 years.
.09 1*30
Solution:
(1 +
)
1
1
.09
1
(1.09) 30  1
= 2000
.09
12.2677
= 2000
= $272,615.08
.09
= 2000
The Sum of an Infinite Geometric
Series
If -1<r<1, then the sum of the infinite geometric series
a1+a1r+a1r2+a1r3+…
in which a1 is the first term and r s the common ration
is given by
a1
S=
1 r
If |r|>1, the infinite series does not have a sum.
Mathematical
Induction
The Principle of Mathematical
Induction
Let Sn be a statement involving the positive
integer n. If
1. S1 is true, and
2. the truth of the statement Sk implies the
truth of the statement S k+1, for every
positive integer k,
then the statement Sn is true for all positive
integers n.
The Steps in a Proof by
Mathematical Induction
Let Sn be a statement involving the positive
integer n. To prove that Sn is true for all
positive integers n requires two steps.
STEP 1 Show that S1 is true
STEP 2 Show that if Sk is assumed to be
true, then Sk+1 is also true, for every positive
integer k.
Example
• For the given statement Sn, write the three statements S1,
Sk, and Sk+1
n(n + 5)
S n = 3 + 4 + 5 + ... + (n + 2) =
2
Solution:
n(n + 5)
S n = 3 + 4 + 5 + ... + (n + 2) =
2
1(1 + 5)
S1 =
=3
2
k (k + 5)
Sk =
2
(k + 1)(( k + 1) + 5) (k + 1)( k + 6)
S k +1 =
=
2
2
Text Example
For the given statement Sn, write the three statements S1, Sk, and Sk+1
.
n(n + 1)
Sn :1 + 2 + 3 + ... + n =
2
Solution We begin with
n(n + 1)
Sn :1 + 2 + 3 + ... + n =
2
Write S1 by taking the first term on the left and replacing n with 1 on the right.
1(1 + 1)
S1 :1 =
2
Text Example cont.
Solution
Write Sk by taking the sum of the first k terms on the left and replacing n with k
on the right.
k(k + 1)
Sk :1 + 2 + 3 + ... + k =
2
Write Sk+1 by taking the sum of the first k + 1 terms on the left and replacing n
with k + 1 on the right.
(k + 1)[(k + 1) + 1]
Sk +1 :1 + 2 + 3 + ... + (k + 1) =
2
(k + 1)(k + 2)
Sk +1 :1 + 2 + 3 + ... + (k + 1) =
2
Simplify on the right.
Example
• Use mathematical induction to prove that
n(n + 5)
3 + 4 + 5 + ... + (n + 2) =
2
Solution:
Step 1:
S n = 3 + 4 + 5 + ... + (n + 2) =
S1 =
1(1 + 5)
=3
2
n(n + 5)
2
Example cont.
• Use mathematical induction to prove that
n(n + 5)
3 + 4 + 5 + ... + (n + 2) =
2
Solution:
Sk =
k (k + 5)
2
k (k + 5)
+ (( k + 1) + 2)
2
k (k + 5) 2(k + 3) k 2 + 5k + 2k + 6
=
+
=
2
2
2
k 2 + 7k + 6 (k + 1)( k + 6)
=
=
2
2
(k + 1)(( k + 1) + 5)
=
2
S k +1 = S k + (( k + 1) + 2) =
Step 2:
The Binomial
Theorem
Patterns in Binomial Expansions
By studying the expanded form of each binomial expression, we are able to
discover the following patterns in the resulting polynomials.
1. The first term is an. The exponent on a decreases by 1 in each successive
term.
2. The exponents on b increase by 1 in each successive term. In the first term,
the exponent on b is 0. (Because b0 = 1, b is not shown in the first term.)
The last term is bn.
3. The sum of the exponents on the variables in any term is equal to n, the
exponent on (a + b)n.
4. There is one more term in the polynomial expansion than there is in the
power of the binomial, n. There are n + 1 terms in the expanded form of
(a + b)n.
Using these observations, the variable parts of the expansion (a + b)6 are
a6,
a5b,
a4b2,
a3b3,
a2b4,
ab5,
b6.
Patterns in Binomial Expansions
Let's now establish a pattern for the coefficients of the terms in the binomial
expansion. Notice that each row in the figure begins and ends with 1. Any
other number in the row can be obtained by adding the two numbers
immediately above it.
Coefficients for (a + b)1.
Coefficients for (a + b)2.
•
•
Coefficients for (a + b) .
3
Coefficients for (a + b) .
4
Coefficients for (a + b)5.
Coefficients for (a + b)6.
•
•
1
2
3
1
3
1
4
6 4 1
• 5 10 10 5 1
1 6 15 20 15 6 1
The above triangular array of coefficients is called Pascal’s triangle. We can
use the numbers in the sixth row and the variable parts we found to write the
expansion for (a + b)6. It is
(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
Definition of a Binomial
n
 
Coefficient  r  .
For nonnegative integers n and r, with
n


n > r, the expression  is called a
r 
 
binomial coefficient and is defined by
n
n!
  =
 r  r !(n  r )!
Example
• Evaluate
Solution:
7
 
3 
n
n!
  =
 r  r !(n  r )!
7
7!
7!
  =
=
 3  3!(7  3)! 3!4!
7 * 6 * 5 * 4 * 3 * 2 *1
=
= 35
3 * 2 *1* 4 * 3 * 2 *1
A Formula for Expanding
Binomials: The Binomial Theorem
• For any positive integer n,
 n  n  n  n 1  n  n 2 2
 n n
(a + b) =  a +  a b +  a b + ... +  b
0
1 
3 
 n
n
Example
• Expand
Solution:
( x + 4)3
( x + 4)3
 3  3  3 2
3 
 3 3
2
=   x +   x * 4 +   x * 4 +  4
 0
1 
 2
 3
Example cont.
• Expand
Solution:
( x + 4)3
( x + 4) 3
 3  3  3 2
3
 3 3
2
=   x +   x * 4 +   x * 4 +  4
 0
1 
 2
 3
3! 3 3! 2 3!
3!
=
x +
4x +
16 x +
64
0!3!
1!2!
2!1!
3!0!
= x 3 + 12 x 2 + 48 x + 64
Finding a Particular Term in a
Binomial Expansion
The rth term of the expansion of
(a+b)n is
 n  n  r +1 r 1

a
b
 r  1
Example
Find the third term in the expansion of (4x-2y)8
Solution:
(4x-2y)8 n=8, r=3, a=4x, b=-2y
 n  n  r +1 r 1
a

b
 r  1
8 
(4 x)83+1 (2 y )31
= 
 3  1
8!
(4 x) 6 (2 y ) 2
=
2!6!
= 28(4096) x 6 4 y 2 = 458,752 x 6 y 2
Text Example
Find the fourth term in the expansion of (3x + 2y)7.
Solution We will use the formula for the rth term of the expansion (a + b)n,
 7
 7
7!
7 3
3
4
3
  (3x) (2y) =   (3x)4 (2y) 3 =
(3x) (2y)
 3
 3
3!(7  3)!
to find the fourth term of (3x + 2y)7. For the fourth term of (3x + 2y)7,
n = 7, r = 4, a = 3x, and b = 2y. Thus, the fourth term is
7!
7 6  5 4!
4
3
4
3
4
3
4 3
(81x )(8y ) =
(81x )(8y ) = 35(81x )(8y ) = 22,680x y
3!4!
3 2 1 4!
Counting Principles,
Permutations, and
Combinations
The Fundamental Counting
Principle
• The number of ways a series of successive
things can occur is found by multiplying the
number of ways in which each thing can
occur.
Example
• Sue goes shopping for an outfit. She buys five blouses,
three skirts and two pairs of shoes. How many different
outfits can she make?
Solution:
(5)(3)(2) = 30
Example
• An executive council is to be formed from a pool of 10
qualified candidates. The council will be composed of a
Chairman, Vice-Chairman, Secretary and Treasurer. All 10
candidates can be appointed for any positions.
Solution:
(10)(9)(8)(7) = 5040
Text Example
Telephone numbers in the United States begin with three-digit area codes
followed by seven-digit local telephone numbers. Area codes and local
telephone numbers cannot begin with 0 or 1. How many different telephone
numbers are possible?
Solution This situation involves making choices with ten groups of items.
Area Code
You cannot use 0 or 1
in these groups. There
are only 8 choices: 2,
3, 4, 5, 6, 7, 8, or 9.
Local Telephone Number
You can use 0, 1, 2, 3, 4, 5, 6, 7, 8,
or 9 in these groups. There are 10
choices per group.
Text Example cont.
Telephone numbers in the United States begin with three-digit area codes
followed by seven-digit local telephone numbers. Area codes and local
telephone numbers cannot begin with 0 or 1. How many different telephone
numbers are possible?
Solution We use the Fundamental Counting Principle to determine the number
of different telephone numbers that are possible. The total number of telephone
numbers possible is
8 2 10 2 10 2 8 2 10 2 10 2 10 2 10 2 10 2 10 = 6,400,000,000.
There are six billion four hundred million different telephone numbers that are
possible.
Permutations
• The number of possible permutations if r
items are taken from n items is
Example
• Given the letters A, B, C, D, E, F and G,
how many arrangements are there of these 7
letters taken 4 at a time.
Solution:
Combinations of n Things Taken
r at a Time
• The number of possible combinations if r
items are taken from n items is
n
n!
  =
 r  r!(n  r )!
Example
• In the game of bridge, each of 4 players is
dealt 13 cards. How many 13-card hands
can be dealt from a 52-card deck?
Solution:
 52 
52!
  =
=
13  13!(52  13)!
52 * 51* 50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * 42 * 41* 40 * 39!
=
13!39!
6.35 1011
Probability
Computing Empirical Probability
The empirical probability of event E is
observed number of times E occurs
P( E ) =
total number of observed occurrences
Example
Hours of Sleep Number of Americans, in millions
4 or less
11
5
24.75
6
68.75
7
82.5
8
74.25
9
8.25
10 or more
5.5
Total 275
An American is randomly selected. Find the
probability of that person getting 6 hours sleep on
a typical night.
Example cont.
Solution:
P( six hours sleep ) =
number of Americans who sleep 6 hours
total number of Americans
68.75 275
=
=
= .25
275 1100
The empirical probability of randomly
selecting an American who gets eight hours
sleep in a typical night is 275/1100 or .25
Computing Theoretical
Probability
If an event E has n(E) equally-likely outcomes and its
Sample space S has n(S) equally-likely outcomes, the
Theoretical probability of event E, denoted by P(E), is
number of outcomes in event E
P( E ) =
number of outcomes in sample space S
The sum of the theoretical probabilities of all possible
Outcomes in the sample is 1.
Text Example
A die is rolled. Find the probability of getting a number less than 5.
Solution The sample space of equally likely outcomes is S = {1, 2, 3,
4, 5, 6}. There are six outcomes in the sample space, so n(S) = 6.
We are interested in the probability of getting a number less
than 5. The event of getting a number less than 5 can be represented by
E = {1, 2, 3, 4}.
There are four outcomes in this event, so n(E) = 4.
The probability of rolling a number less than 5 is
n(E) 4 2
P(E) =
= =
n(S) 6 3
Example
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
Example cont.
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
Example cont.
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution:
E = {TTT , THT , HTT , TTH , HHT , HTH , THH }
n( E ) = 7
S = {TTT , THT , HTT , TTH , HHT , HTH , THH , HHH }
n( S ) = 8
n( E ) 7
P( E ) =
=
n( S ) 8
The probability of getting at most 2 heads
when a coin is tossed 3 times is 7/8
The Probability of an Event Not
Occurring
• The probability that an event E will not
occur is equal to one minus the probability
that it will occur.
P(not E) = 1 - P(E)
Or Probabilities with Mutually
Exclusive Events
If A and B are mutually exclusive events,
then
P(A or B) = P(A) + P(B).
Text Example
If one card is randomly selected from a deck of cards, what is the probability
of selecting a king or a queen?
Solution We find the probability that either of these mutually
exclusive events will occur by adding their individual probabilities.
4
4
8
2
+
=
=
P(king or queen) = P(king) + P(queen) =
52 52 52 13
The probability of selecting a king or a queen is 2/13 .
Or Probabilities with Events That
Are Not Mutually Exclusive
• If A and B are not mutually exclusive
events, then
• P(A or B) = P(A) + P(B) – P(A and B).
Text Example
The figure illustrates a spinner. It is
equally probable that the pointer will
land on any one of the eight regions,
numbered 1 through 8. If the pointer
lands on a borderline, spin again. Find
the probability that the pointer will
stop on an even number or a number
greater than 5.
Text Example cont.
Solution It is possible for the pointer to land on a number that is even
and greater than 5. Two of the numbers, 6 and 8, are even and greater
than 5. These events are not mutually exclusive. The probability of
landing on a number that is even and greater than 5 is
 even or

 even and

 = P(even) + P(greater than 5)  P

P
greater
than
5
greater
than
5




4
=
8
Four of the eight numbers, 2,
4, 6, and 8, are even.
+
3
8

2
8
Three of the eight numbers, 6, 7,
and 8, are greater than 5.
4 +32 5
=
=
8
8
Two of the eight numbers, 6 and 8,
are even and greater than 5.
The probability that the pointer will stop on an even number or on a number
greater that 5 is 5/8.
And Probabilities with
Independent Events
• If events A and B are Independent, then the
probability of A and B is simply:
P(A and B) = P(A) · P(B)