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.

Write the first four terms of the sequence whose nth term, or general term, is given: an 3n 4.

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, ….

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.

3 1 4 3 4 7a1, 1st term 3 2 4 6 4

10

a2, 2nd term

3 3 4 9 4 13

a3, 3rd term 3 4 4 12 4 16a4, 4th

term

Text Example

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

Where i is the index of summation, n is the upper limit of summation,

and 1 is the lower limit of summation.

n

n

i i aaaaaa ...43211

Example

5

1

)25(i

i• Expand and evaluate the sum:

Solution:

5

53113

)25(5

1

i

i

Example

16

1

8

1

4

1

2

11

• Express the sum using summation notation:

Solution:

5

112

116

1

8

1

4

1

2

11

ii

Example

12

1...

8

1

4

1

2

11

n

• Express the sum using summation notation:

Solution:

n

ii

n

11

1

2

12

1...

8

1

4

1

2

11

Arithmetic Sequences

A mathematical model for the average annual salaries of major league baseball players generates the following data.

1,438,0001,347,0001,256,0001,165,0001,074,000983,000892,00

0801,000Salary

19981997199619951994199319921991Year

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.

Arithmetic Sequences

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.

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

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.

The first five terms are

624, 600, 576, 552, and 528.

Solution

a1 624 This is given. a2 a1 – 24 624 – 24 600 Use an an 1 24 with n 2.

a3 a2 – 24 600 – 24 576 Use an an 1 24 with n 3.

a4 a3 – 24 576 – 24 552 Use an an 1 24 with n 4.

a5 a4 – 24 552 – 24 528 Use an an 1 24 with n 5.

Text Example cont.

Example• Write the first six terms of the arithmetic sequence where

a1 = 50 and d = 22

•a1 = 50 a2 = 72 a3 = 94 a4 = 116 a5 = 138 a6 = 160

Solution:

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

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(8 1)(7) 4 ) 4 (49) 5

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.

Text Example

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

in which a1 is the first term and an is the nth term.

)(2 1 nn aan

S

Example

690)69(10

)636(10

)3*196(6(10

)(2

2020120

aaS

Solution:

• Find the sum of the first 20 terms of the arithmetic sequence: 6, 9, 12, 15, ...

Example

15

1

)42(i

i

18012*15

)24(2

15

))26()2((2

15

)42(15

1

i

i

• Find the indicated sum

Solution:

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 Common ratio1, 5 25, 125, 625... r 5 1 = 56, -12, 24, -48, 96.... r -12 6 -2

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 1/3, or 2. The third term is 21/3, or2/3.The fourth term is 2/31/3, or2/9, and so on. The first six terms are

6, 2,2/3,2/9,2/27,2/81.

Text Example

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

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

a88 1 4)7 4(128)

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.

Text Example

The Sum of the First n Terms of a Geometric Sequence

r

raS

n

n

1

)1(1

The sum, Sn, of the first n terms of a geometric sequence is given by

in which a1 is the first term and r is the common ratio.

Example

5314404

)5314411(4

)3(1

))3(1(4

1

)1(

1

)1(

12121

12

1

r

raS

r

raS

n

n

• Find the sum of the first 12 terms of the geometric sequence: 4, -12, 36, -108, ...Solution:

Value of an Annuity: Interest Compounded n Times per YearIf 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:

nr

nr

PA

nt 1)1(

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

109.

1)109.

1(2000

1)1( 30*1

nr

nr

PA

nt

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.

Solution:

08.615,272$09.

2677.122000

09.

1)09.1(2000

1

09.

1)1

09.1(

2000

30

30*1

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

r

aS

11

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

2

)5()2(...543

nnnSn

• For the given statement Sn, write the three statements S1, Sk, and Sk+1

2

)6)(1(

2

)5)1)((1(2

)5(

32

)51(12

)5()2(...543

1

1

kkkk

S

kkS

S

nnnS

k

k

n

Solution:

For the given statement Sn, write the three statements S1, Sk, and Sk+1

.

Text Example

Solution We begin with

Sn:1 2 3 ... n n(n 1)

2

Sn:1 2 3 ... n n(n 1)

2

Write S1 by taking the first term on the left and replacing n with 1 on the right.

S1:1 1(1 1)

2

Solution

Write Sk by taking the sum of the first k terms on the left and replacing n with k on the right.

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.

Simplify on the right.

Text Example cont.

Sk:1 2 3 ... k k(k 1)

2

Sk1:1 2 3 ... (k 1) (k 1)[(k 1) 1]

2

Sk1:1 2 3 ... (k 1) (k 1)(k 2)

2

Example

2

)5()2(...543

nnn

• Use mathematical induction to prove that

Solution:

32

)51(12

)5()2(...543

1

S

nnnSn

Step 1:

Example cont.

2

)5()2(...543

nnn

• Use mathematical induction to prove that

Solution:

Step 2:

2

)5)1)((1(2

)6)(1(

2

67

2

625

2

)3(2

2

)5(

)2)1((2

)5()2)1((

2

)5(

2

2

1

kk

kkkk

kkkkkk

kkk

kSS

kkS

kk

k

The Binomial Theorem

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 1

• 3 3 1• 4 6 4 1

• 5 10 10 5 11 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

Patterns in Binomial Expansions

Definition of a Binomial

Coefficient .For nonnegative integers n and r, with

n > r, the expression is called a

binomial coefficient and is defined by

r

n

r

n

)!(!

!

rnr

n

r

n

Example

3

7• Evaluate

Solution:

351*2*3*4*1*2*3

1*2*3*4*5*6*7

!4!3

!7

)!37(!3

!7

3

7

)!(!

!

rnr

n

r

n

A Formula for Expanding Binomials: The Binomial Theorem

nnnnn bn

nba

nba

na

nba

...

310)( 221

• For any positive integer n,

Example3)4( x• Expand

Solution:

3223

3

43

34*

2

34*

1

3

0

3

)4(

xxx

x

Example cont.3)4( x• Expand

Solution:

644812

64!0!3

!316

!1!2

!34

!2!1

!3

!3!0

!3

43

34*

2

34*

1

3

0

3

)4(

23

23

3223

3

xxx

xxx

xxx

x

Finding a Particular Term in a Binomial Expansion

The rth term of the expansion of (a+b)n is

11

1

rrn bar

n

ExampleFind the third term in the expansion of (4x-2y)8

(4x-2y)8 n=8, r=3, a=4x, b=-2ySolution:

2626

26

13138

11

752,4584)4096(28

)2()4(!6!2

!8

)2()4(13

8

1

yxyx

yx

yx

bar

n rrn

Find the fourth term in the expansion of (3x 2y)7.

Text Example

7

3

(3x)7 3(2y)3

7

3

(3x)4 (2y)3

7!

3!(7 3)!(3x)4 (2y)3

Solution We will use the formula for the rth term of the expansion (a b)n,

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!

3!4!(81x4 )(8y3 )

7 6 54!

32 14!(81x 4 )(8y3) 35(81x 4)(8y3) 22,680x 4y3

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?

(5)(3)(2) = 30Solution:

• An executive council is to be formed from a pool of 10 qualified candidates. The council will be composed of aChairman, Vice-Chairman, Secretary and Treasurer. All 10 candidates can be appointed for any positions.

(10)(9)(8)(7) = 5040Solution:

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?

Area Code Local Telephone Number

Solution This situation involves making choices with ten groups of items.

You cannot use 0 or 1 in these groups. There are only 8 choices: 2, 3, 4, 5, 6, 7, 8, or 9.

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

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 10 10 8 10 10 10 10 10 10 = 6,400,000,000.

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?

There are six billion four hundred million different telephone numbers that are possible.

Text Example cont.

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

)!(!

!

rnr

n

r

n

• The number of possible combinations if r items are taken from n items is

111035.6

!39!13

!39*40*41*42*43*44*45*46*47*48*49*50*51*52

)!1352(!13

!52

13

52

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:

Probability

soccurrenceobservedofnumbertotal

occursEtimesofnumberobservedEP )(

Computing Empirical Probability

The empirical probability of event E is

Example

An American is randomly selected. Find theprobability of that person getting 6 hours sleep on

a typical night.

Hours of Sleep Number of Americans, in millions4 or less 115 24.756 68.757 82.58 74.259 8.2510 or more 5.5

Total 275

Example cont.

The empirical probability of randomly

selecting an American who gets eight hours

sleep in a typical night is 275/1100 or .25

Americansofnumbertotal

hourssleepwhoAmericansofnumbersleephourssixP

6)(

25.1100

275

275

75.68

Solution:

Computing Theoretical Probability

Sspacesampleinoutcomesofnumber

EeventinoutcomesofnumberEP )(

If an event E has n(E) equally-likely outcomes and itsSample space S has n(S) equally-likely outcomes, the Theoretical probability of event E, denoted by P(E), is

The sum of the theoretical probabilities of all possibleOutcomes in the sample is 1.

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

Text Example

P(E) n(E)

n(S)

4

6

2

3

ExampleWhat 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:

8

7

)(

)()(

8)(

},,,,,,,{

7)(

},,,,,,{

Sn

EnEP

Sn

HHHTHHHTHHHTTTHHTTTHTTTTS

En

THHHTHHHTTTHHTTTHTTTTE

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).

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.

P(king or queen) P(king) P(queen)

The probability of selecting a king or a queen is 2/13 .

4

52

4

52

8

52

2

13

Text Example

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).

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

The probability that the pointer will stop on an even number or on a number greater that 5 is 5/

8.

Two of the eight numbers, 6 and 8, are

even and greater than 5.

Four of the eight numbers, 2, 4, 6, and 8, are even.

Three of the eight numbers, 6, 7, and 8, are 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

Peven or

greater than 5

P(even) P(greater than 5) P

even and

greater than 5

4

8

3

8

2

8

4 3 2

8

5

8

P(A and B) = P(A) · P(B)

And Probabilities with Independent Events

• If events A and B are Independent, then the probability of A and B is simply: