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. 3 · 1 + 4 = 3 + 4 = 7 a1, 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 = 16 a4, 4th term 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 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: Solution:

Example Express the sum using summation notation: Solution:

Example Express the sum using summation notation: Solution:

Arithmetic Sequences

Arithmetic Sequences A mathematical model for the average annual salaries of major league baseball players generates the following data. 1,438,000 1,347,000 1,256,000 1,165,000 1,074,000 983,000 892,000 801,000 Salary 1998 1997 1996 1995 1994 1993 1992 1991 Year 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 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. 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 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:

Example 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 ratio 1, 5 25, 125, 625... r = 5 ¸ 1 = 5 6, -12, 24, -48, 96.... 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  1/3, or 2. The third term is 2  1/3, or 2/3. The fourth term is 2/3  1/3, or 2/9, and so on. The first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81.

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 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:

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:

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

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:

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 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 S1 is true, and 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 Solution:

Text Example For the given statement Sn, write the three statements S1, Sk, and Sk+1 . Solution We begin with Write S1 by taking the first term on the left and replacing n with 1 on the right.

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. 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.

Example Use mathematical induction to prove that Solution: Step 1:

Example cont. Use mathematical induction to prove that Solution: 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 1 3 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 Coefficient . For nonnegative integers n and r, with n > r, the expression is called a binomial coefficient and is defined by

Example Evaluate Solution:

A Formula for Expanding Binomials: The Binomial Theorem For any positive integer n,

Example Expand Solution:

Example cont. Expand Solution:

Finding a Particular Term in a Binomial Expansion The rth term of the expansion of (a+b)n is

Example Find the third term in the expansion of (4x-2y)8 Solution: (4x-2y)8 n=8, r=3, a=4x, b=-2y

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, 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

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? 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 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

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

Computing Empirical Probability The empirical probability of event E is

Example An American is randomly selected. Find the probability of that person getting 6 hours sleep on a typical night.

Example cont. The empirical probability of randomly Solution: 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 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

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: 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. P(king or queen) = P(king) + P(queen) = 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 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. 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)