# Chapter 8 power point

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```					Counting without Counting
CHAPTER 8
Combinatorics
and Probability.

Arithmetic
Sequences and
Series
An infinite sequence is a function whose domain
is the set of positive integers.
a1, a2, a3, a4, . . . , an, . . .

terms

The first three terms of the sequence an = 4n – 7 are
a1 = 4(1) – 7 = – 3
a2 = 4(2) – 7 = 1              finite sequence
a3 = 4(3) – 7 = 5.
A sequence is arithmetic if the differences
between consecutive terms are the same.

4, 9, 14, 19, 24, . . .
arithmetic sequence
9–4=5
14 – 9 = 5
The common difference, d, is 5.
19 – 14 = 5
24 – 19 = 5
Example: Find the first five terms of the sequence
and determine if it is arithmetic.
an = 1 + (n – 1)4
a1 = 1 + (1 – 1)4 = 1 + 0 = 1
a2 = 1 + (2 – 1)4 = 1 + 4 = 5
a3 = 1 + (3 – 1)4 = 1 + 8 = 9        d=4
a4 = 1 + (4 – 1)4 = 1 + 12 = 13
a5 = 1 + (5 – 1)4 = 1 + 16 = 17

This is an arithmetic sequence.
The nth term of an arithmetic sequence has the
form
an = dn + c
where d is the common difference and c = a1 – d.

a1 = 2                                 c=2–6=–4
2, 8, 14, 20, 26, . . . .
d=8–2=6

The nth term is 6n – 4.
Example: Find the formula for the nth term of an
arithmetic sequence whose common difference is 4
and whose first term is 15. Find the first five terms
of the sequence.
an = dn + c
a1 – d = 15 – 4 = 11
= 4n + 11
The first five terms are
a1 = 15
15, 19, 23, 27, 31.
d=4
The sum of a finite arithmetic sequence with n
terms is given by
S n  n (a1  an).
2

5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 = ?

n = 10
a1 = 5                       a10 = 50

Sn  10 (5  50)  5(55)  275
2
The sum of the first n terms of an infinite sequence
is called the nth partial sum.
Sn  n (a1  an)
2
Example: Find the 50th partial sum of the arithmetic
sequence – 6, – 2, 2, 6, . . .

a1 = – 6       d=4         c = a1 – d = – 10
an = dn + c = 4n – 10      a50 = 4(50) – 10 = 190

Sn  50 (6  190)  25(184)  4600
2
The sum of the first n terms of a sequence is
represented by summation notation.
upper limit of summation
n

a  a  a
i 1
i     1   2    a3  a4     an

index of              lower limit of summation
summation
5

 1  n   (1 1)  (1 2)  (1 3)  (1 4)  (1 5)
i 1
 2  3 4 5 6
 20
Example: Find the partial sum.
100

  2n   2(1)  2(2)  2(3)        2(100)
i 1
 246       200

a1                         a100

S100  n (a1  a100)  100 (2  200)
2                2
 50(202)  10,100
Consider the infinite sequence a1, a2, a3, . . ., ai, . . ..

1. The sum of the first n terms of the sequence is called
a finite series or the partial sum of the sequence.
n
a1 + a2 + a3 + . . . + an   ai
i 1

2. The sum of all the terms of the infinite sequence is
called an infinite series.

a1 + a2 + a3 + . . . + ai + . . .   ai
i 1
 
          i
Example: Find the fourth partial sum of  5 1 .
i 1
2

         
4      i         1      2        3           4
 2
i1
5 1  5 1 5 1 5 1 5 1
2    2    2    2

 5 1   5 1   5 1   5 1 
2        4        8       16
 555 5
2 4 8 16
 40  20  10  5  75
16 16 16 16 16
Section 8.3

Geometric
Sequences and
Series
An infinite sequence is a function whose domain
is the set of positive integers.
a1, a2, a3, a4, . . . , an, . . .

terms

The first three terms of the sequence an = 2n2 are
a1 = 2(1)2 = 2
a2 = 2(2)2 = 8             finite sequence
a3 = 2(3)2 = 18.
A sequence is geometric if the ratios of consecutive
terms are the same.
2, 8, 32, 128, 512, . . .
8 4                          geometric sequence
2
32  4
8
The common ratio, r, is 4.
128  4
32
512  4
128
The nth term of a geometric sequence has the form
an = a1rn - 1
where r is the common ratio of consecutive terms of
the sequence.
r  75  5
15
a1 = 15
15, 75, 375, 1875, . . .
a2 =    a3 =   a4 =
15(5)   15(52) 15(53)

The nth term is 15(5n-1).
Example: Find the 9th term of the geometric sequence
7, 21, 63, . . .

a1 = 7        r  21  3
7

an = a1rn – 1 = 7(3)n – 1

a9 = 7(3)9 – 1 = 7(3)8

= 7(6561) = 45,927

The 9th term is 45,927.
The sum of the first n terms of a sequence is
represented by summation notation.
upper limit of summation
n

a  a  a
i 1
i     1   2    a3  a4     an

index of                 lower limit of summation
summation

5

 4  41  42  43  44  45
n

 4 16  64  256 1024
n1

 1364
The sum of a finite geometric sequence is given by

       
n
Sn   a1r    i 1
 a1 1 rn .
i 1
1 r

5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ?

n=8
a1 = 5                        r  10  2
5

    
Sn  a1 1  r n  5 1  28  5 1  256  5 255  1275
1 r        1 2        1 2         
1        
The sum of the terms of an infinite geometric
sequence is called a geometric series.

If |r| < 1, then the infinite geometric series

a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . .

has the sum S   a1r 
a1 i
.
i 0
1 r

If r  1 , then the series does not have a sum.
Example: Find the sum of 3  1  1  1 
3 9
r  1
3

 a1  
S            3
 1 r  1  1

3 
 3  3  3 3  9
1 1 4      4 4
3 3
The sum of the series is 9 .
4
Section 8.4

Mathematical
Induction
Mathematical induction is a legitimate method of
proof for all positive integers n.

Principle:
Let Pn be a statement involving n, a positive integer.
If
1. P1 is true, and
2. the truth of Pk implies the truth of Pk + 1 for
every positive k,
then Pn must be true for all positive integers n.
Example:
3(2k  1)
Find Pk + 1 for Pk : Sk                 .
k 1

3[2(k  1)  1]
Pk 1 : Sk 1                          Replace k by k + 1.
k  1 1
3(2k  2  1)
                           Simplify.
k
3(2k  3)
                           Simplify.
k
Example:
Use mathematical induction to prove
Sn = 2 + 4 + 6 + 8 + . . . + 2n = n(n + 1)
for every positive integer n.

1. Show that the formula is true when n = 1.
S1 = n(n + 1) = 1(1 + 1) = 2      True
2. Assume the formula is valid for some integer k.
Use this assumption to prove the formula is valid
for the next integer, k + 1 and show that the
formula Sk + 1 = (k + 1)(k + 2) is true.
Sk = 2 + 4 + 6 + 8 + . . . + 2k = k(k + 1)    Assumption
Example continues.
Example continued:
Sk + 1 = 2 + 4 + 6 + 8 + . . . + 2k + [2(k + 1)]

= 2 + 4 + 6 + 8 + . . . + 2k + (2k + 2)

= Sk + (2k + 2)              Group terms to form Sk.

= k(k + 1) + (2k + 2)        Replace Sk by k(k + 1).
= k2 + k + 2k + 2            Simplify.
= k2 + 3k + 2

= (k + 1)(k + 2)

= (k + 1)((k + 1)+1)
The formula Sn = n(n + 1) is valid for all positive integer
values of n.
Sums of Powers of Integers :
n
n(n  1)
1.   i  1 2  3  4 
i 1
n
2
n
n(n  1)(2n  1)
2.    i 2  12  22  32  42 
i 1
 n2 
6
n
n 2(n  1) 2
3.    i 3  13  23  33  43 
i 1
 n3 
4
n
n(n  1)(2n  1)(3n 2  3n  1)
4.    i 4  14  24  34  44 
i 1
 n4 
30
n
n 2(n  1) 2(2n 2  2n  1)
5.    i 5  15  25  35  45 
i 1
 n5 
12
Example:
Use mathematical induction to prove for all positive integers n,
n
n(n  1)(2n  1)
 i 2  12  22  32  42 
i 1
 n2 
6
.

11  1)(2(1)  1) 1(2)(2  1) 6
(
S1                                   1               True
6              6        6
k(k  1)(2k  1)
Sk  12  22  32  42   k 2                          Assumption
6
S k 1  12  22  32  42   k 2  (k  1) 2
 S k  (k  1) 2
 S k  k 2  2k  1                         Group terms to form Sk.

k(k  1)(2k  1) 2
                    k  2k  1             Replace Sk by k(k + 1).
6                                       Example continues.
Example continued:

 2k 3  3k 2  k  6k 2  12k  6       Simplify.
6                 6
 2k 3  9k 2  13k  6
6
(k 2  3k  2)(2k  3)

6
(k  1)(k  2)(2k  3)

6
(k  1)[(k  1)  1][2(k  1)  1]

6
The formula Sn  n(n  1)(2n  1) is valid for all positive
6
integer values of n.
Finite Differences
The first differences of the sequence 1, 4, 9, 16, 25, 36 are
found by subtracting consecutive terms.
n:     1        2       3       4        5         6
an:    1        4       9       16       25        36

First differences:         3       5       7        9        11

Second differences:              2       2       2         2

The second differences are found by subtracting consecutive
first differences.
When the second differences are all the same nonzero number,
the sequence has a perfect quadratic model.
Find the quadratic model for the sequence
1, 4, 9, 16, 25, 36, . . .

an = an2 + bn + c
a1 = a(1)2 + b(1) + c = 1            a+ b+c=1
a2 = a(2)2 + b(2) + c = 4          4a + 2b + c = 4
a3 = a(3)2 + b(3) + c = 9          9a + 3b + c = 9

Solving the system yields a = 1, b = 0, and c = 0.
an = n2
Example:
Find the quadratic model for the sequence with
a0 = 3, a1 = 3, a4 = 15.
an = an2 + bn + c
a0 = a(0)2 + b(0) + c = 3
a1 = a(1)2 + b(1) + c = 3
a4 = a(4)2 + b(4) + c = 15

c= 3
a+ b+ c= 3              Solving the system yields
a = 1, b = –1, and c = 3.
16a + 4b + c = 15
an = n2 – n + 3
Section 8.6-8.7

Counting Principles
and Probability
The Fundamental Counting Principle states that
if one event can occur m ways and a second event
can occur n ways, the number of ways the two
events can occur in sequence is m • n.

Start
1st Coin
Tossed
Heads                   Tails           2 ways to flip the coin.
2nd Coin
Tossed

There are 2  2 different outcomes: {HH, HT, TH, TT}.
Example: A meal consists of a main dish, a side
dish, and a dessert. How many different meals can
be selected if there are 4 main dishes, 2 side dishes
and 5 desserts available?

# of              # of             # of
main dishes       side dishes       desserts

4              2               5       =   40

There are 40 meals available.
A permutation is an ordered arrangement of n
different elements.
How many permutations are possible using the three
colors red, white, and blue?
There are 3 choices for the first color, 2 choices for the
second color and only 1 choice for the third color.

3! = 3 • 2 • 1
= 6 permutations

“factorial”
A permutation of n elements taken r at a time is
a subset of the collection of elements where order is
important.
Five projects are entered in a science contest. In
how many ways can the projects come in first,
second, and third?

1st    2nd    3rd         3 projects
5    4    3
__ x __ x __
5 projects
4 projects

5 • 4 • 3 = 60 ways
The formula for the number of permutations of n
elements taken r at a time is
n! .
n Pr    (n  r)!
# in the
collection
# taken
from the
collection

Pr  5 P3  5!  5  4  3  2 1  60
n
2!        2 1
If some of the items are identical, distinguishable
permutations must be used.

In how many distinguishable ways can the letters
STATS be written?

STATS

The T’s are not distinguishable.
S’s

Example continues.
The number of distinguishable permutations of
the n objects is
n!
n1!  n2 !  n3 !  nk !
where n = n1 + n2 + n3 + . . . + nk.

The letters STATS can be written in
5!       120  30 ways.
2!  2! 1!    4

S’s       T’s    A’s
A combination of n elements taken r at a time is
a subset of the collection of elements where order is
not important.
Using the letters A, B, C, and D, find all the
possible combinations using two of the letters.
{AB}             This is the same as {BA}.
{AC}
{BC}                   combinations using 2 of
the 4 letters.
{BD}
{CD}
The formula for the number of combinations of n
elements taken r at a time is
n! .
nC r    (n  r)!r !
# in the
collection
# taken
from the
collection

C r  4 C 2  4!  4  3  2 1  6
n
2!2! 2 1  2 1
Example: How many different ways are there to
choose 6 out of 10 books if the order does not
matter?

3
C r  10 C 6  10!  10  9  8  7  6!  210
n
4!6! 4  3  2 1  6!

There are 210 ways to choose the 6 books.

```
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