Chapter 8 Review and Tests

Review

Definitions, Concepts, and Formulas Examples

8.1 Sequences and Series

  1. An infinite sequence is a function whose domain is the set of positive integers. The function values a1, a2, a3, , an, are called the terms of the sequence. The term an is the nth term, or general term, of the sequence.

  2. Recursive formula

    A sequence may be defined recursively, with the nth term of the sequence defined in relation to previous terms.

  3. Factorial notation

    n!=n(n1)  321 and 0!=1
  4. Summation notation

    i=1nai=a1+a2+a3++an
  5. Series

    Given an infinite sequence a1, a2, a3,, an,, the sum i=1nai=a1+a2+a3+ is called a series and the sum i=1nai is called the nth partial sum of the series.

  1. The infinite sequence 1, 14, 19, 116,  ,1n2,  has an=1n2 as its general term.

    The associated series sums these terms, 1+14+19+116+  +1n2+ 

  2. The sequence a1=4, ak=3ak1 (k2) is defined recursively.

    The first five terms of this sequence are 4,12,36, 108, 324.

  3. 7!=7654321

  4. i=11i2 is the summation notation for the series associated with the sequence with general term an=1n2.

  5. i=141i2=1+14+19+116 is a partial sum of the series i=11i2.

8.2 Arithmetic Sequences; Partial Sums

  1. A sequence is an arithmetic sequence if each term after the first differs from the preceding term by a constant. The constant difference d between the consecutive terms is called the common difference. We have d=anan1 for all n2.

  2. The nth term an of the arithmetic sequence a1, a2, a3,, an, is given by an=a1+(n1)d, n1 where d is the common difference.

  3. The sum Sn of the first n terms of the arithmetic sequence a1, a2, , an,  is given by the formula

    Sn=n(a1+an2)=n(2a1+(n1)d2).
  1. The arithmetic sequence 3, 5, 7, 9, 11, has

    a1=3andd=53=2Common difference

    Any other two consecutive terms could be used to find this ­common difference.

  2. We can use the general term an=a1+(n1)d to find the 20th term of this sequence.

    a20=3+(201)2a1=3, d=2a20=3+192=41
  3. The sum of the first 20 terms of this sequence is

    S20=20 (3+412)=20(22)=440
    .

8.3 Geometric Sequences and Series

  1. A sequence is a geometric sequence if each term after the first is a constant multiple of the preceding term. The constant ratio r between the consecutive terms is called the common ratio. We have an+1an=r, n1.

  2. The nth term an of the geometric sequence a1, a2,  , an,  is given by an=a1rn1, n1, where r is the common ratio.

  3. The sum Sn of the first n term of the geometric sequence a1, a2,, an, with common ratio r1 is given by the formula

    Sn=a1(1rn)1r.

    The sum S of an infinite geometric sequence is given by

    S=i=1a1ri1=a11r if |r|<1.
  4. An annuity is a sequence of equal payments made at equal time intervals. Suppose $P is the payment made at the end of each of n compounding periods year and i is the annual interest rate. Then the value A of the annuity after t year is

    A=P(1+in)nt1in.
  5. If |r|<1, the infinite geometric series a1+a1r+a1r2+  +a1rn1+ has the sum S=a11r. When |r|1, the series does not have a finite sum.

  1. The geometric sequence 12, 14, 18, 116, has a1=12 and r=14÷12=12. Any other two consecutive terms could be used to find r.

  2. The sixth term a6=12(12)61=(12)6=164

  3. The sum of the first six terms is S6=12[1(12)6]112=12[1(12)6]12=1(12)6=6364.

    The sum of the terms of this infinite geometric sequence is

    S=12112=1212=1.
  4. The value of an annuity in which $100 is deposited at the end of each month, earning 6% interest compounded monthly for 20 years (to the nearest dollar), is

    A=100[(1+0.0612)12201]0.0612=$46,204.

8.4 Mathematical Induction

The statement Pn is true for all positive integers n if the following properties hold:

  1. P1 is true.

  2. If Pk is a true statement, then Pk+1 is a true statement.

See Examples 2 and 3 in Section 8.4.

8.5 The Binomial Theorem

  1. Binomial coefficient: (nj)=n!j!(nj)!, 0jn

    (n0)=1=(nn)
  2. Binomial Theorem:

    (x+y)n=(n0)xn+(n1)xn1y+(n2)xn2y2++(nj)xnjyj+  +(nn)yn=j=0n(nj)xnjyj
  3. The term containing the factor xr in the expansion of (x+y)n is

    (nnr)xrynr.
  1. (75)=7!5!(75)!=7!5! 2!=76543215432121=7621=21

  2. (2x3)4=(2x)4+4!1! 3!(2x)3(3)+4!2! 2!(2x)2(3)2+4!3!1!(2x)(3)3+(3)4=24x4+4(2)3x3(3)+6(2)2x2(9)+4(2)x(27)+81=16x496x3+216x2216x+81

  3. The term containing the factor x3 in the expansion of (2x+y)7 is

    (773)(2x)3y73=280x3y4.

8.6 Counting Principles

  1. Fundamental Counting Principle: If a first choice can be made in p different ways, a second choice in q different ways, a third choice in r different ways, and so on, then the sequence of choices can be made in pqr different ways.

  2. A permutation is an arrangement of n distinct objects in a fixed order in which no object is used more than once. The order in an arrangement is important.

  3. Permutation formula: The number of permutations of n distinct objects taken r at a time is

    P(n,r)=n(n1)(n2)  (nr+1)=n!(nr)!
    .
  4. Combination formula: When r objects are chosen from n distinct objects without regard to order, we call the set of r objects a combination of n objects taken r at a time. The number of combinations of n distinct objects taken r at a time is denoted by C(n, r), where

    C(n,r)=n!(nr)!r!.
  5. Distinguishable permutations: The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is

    n!n1!n2!nk!,

    where n1+n2+  +nk=n.

  1. By the Fundamental Counting Principle, if there are 5 choices for a main dish and 3 choices for a side dish, then there are 53=15 choices for a meal consisting of one main dish and one side dish.

  1. The number of ways you can arrange a playlist of 4 songs chosen from 9 songs is found using the formula for the number of ­permutations of 9 objects taken 4 at a time.

    P(9, 4)=9!(94)!=9!5!=3,024
  2. If you are not determining the order in which the songs will be played when choosing 4 songs from 9 songs, this is a combination. This can be done in

    C(9, 4)=9!(94)!4!=9!5! 4!=126 ways.
  3. Given 3 cards with 1, 2, 3 written on the front (one number on each card) and D, A, D written on the back (one letter on each card), the number of ways the front of the cards can be arranged is 3!=6, the number of permutations of 3 objects.

    However, the backs of the cards during these permutations show the number of distinguishable permutations of 3 objects where 2 are of one kind (the letter D) and 1 is of a second kind (the letter A). There are 3!2! 1!=3 such permutations, one for each position in which D can occur.

8.7 Probability

  1. Experiment: Any process that terminates in one or more outcomes (results) is an experiment.

  2. Sample space: The set of all possible outcomes of an ­experiment is called the sample space of the experiment.

  3. Event: An event is any subset of a sample space.

  4. Equally likely events: Outcomes of an experiment are said to be equality likely if no outcome should result more often than any other outcome when the experiment is performed repeatedly.

  5. Probability: If all of the outcomes in a sample space S are equally likely, the probability of an event E, denoted P(E), is defined by

    P(E)=n(E)n(S),

    where n(E) is the number of outcomes in E and n(S) is the total number of outcomes in S.

    1. The Additive Rule: If E and F are events in a sample space S, then

      P(E or F)=P(EF)=P(E)+P(F)P(EF).
    2. If EF=, E and F are called mutually exclusive events. For mutually exclusive events E and F, P(EF)=P(E)+P(F).

    3. P(E)=P(not E)=1P(E); the event E is called the complement of the event E.

  1. i-iv. Our experiment consists of tossing a single fair die so that our sample space is S={1, 2, 3, 4, 5, 6}. To find the probability that a number greater than 2 occurs, we find the probability of the event, E={3, 4, 5, 6}. When tossing a fair die, each number is as likely as any other to occur.

  2. v. The probability of E is found as

    P(E)=n(E)n(S)=46=23
    .
    1. The events E={3, 4, 5, 6} and F={1} are mutually ­exclusive (EF=) So the probability that a number greater than 2 occurs or a 1 occurs is

      P(EF)=P(E)+P(F)=23+16=56.P(E)=P(not E)=1P(E)=123=13.
    1. The event E={1, 2}.

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