Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

1



Supporting Australian Mathematics Project

2


3


4


5




6



7

A guide for teachers – Years 11 and 12
Algebra and coordinate geometry: Module 6

The binomial theorem



8

9

10

11

12

The bionomial theorem – A guide for teachers (Years 11–12)
Principal author: Dr Michael Evans AMSI
Peter Brown, University of NSW
Associate Professor David Hunt, University of NSW
Dr Daniel Mathews, Monash University
Editor: Dr Jane Pitkethly, La Trobe University
Illustrations and web design: Catherine Tan, Michael Shaw
Full bibliographic details are available from Education Services Australia
...
edu
...
esa
...
au
© 2013 Education Services Australia Ltd, except where indicated otherwise
...

This publication is funded by the Australian Government Department of Education,
Employment and Workplace Relations
...
org
...
amsi
...
au

Assumed knowledge
...


4

Content
...


6

Expansions and the notation

n
r


...
13
Pascal’s triangle — the observations
...
20
Proof of the binomial theorem by mathematical induction
...
25

Links forward
...
30
Series for e
...
31

History and applications
...
31
History
...
33

The binomial
theorem

Assumed knowledge
• Basic skills for simplifying algebraic expressions
...

• Factoring linear and quadratic expressions
...


Motivation
I am the very model of a modern Major-General,
I’ve information vegetable, animal, and mineral,
I know the kings of England, and I quote the fights historical
From Marathon to Waterloo, in order categorical;
I’m very well acquainted, too, with matters mathematical,
I understand equations, both the simple and quadratical,
About binomial theorem I’m teeming with a lot o’ news —
With many cheerful facts about the square of the hypotenuse
...


When you look at the following expansions you can see the symmetry and the emerging
patterns
...

(a + b)2 = (a + b)(a + b)
= a 2 + 2ab + b 2

A guide for teachers – Years 11 and 12

• {5}

(a + b)3 = (a + b)2 (a + b)
= (a 2 + 2ab + b 2 )(a + b)
= (a 3 + 2a 2 b + ab 2 ) + (a 2 b + 2ab 2 + b 3 )
= a 3 + 3a 2 b + 3ab 2 + b 3
(a + b)4 = (a + b)3 (a + b)
= (a 3 + 3a 2 b + 3ab 2 + b 3 )(a + b)
= (a 4 + 3a 3 b + 3a 2 b 2 + ab 3 ) + (a 3 b + 3a 2 b 2 + 3ab 3 + b 4 )
= a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4
Notice that

• the expansion of (a + b)2 has three terms and in each term the sum of the indices is 2
• the expansion of (a + b)3 has four terms and in each term the sum of the indices is 3
• the expansion of (a + b)4 has five terms and in each term the sum of the indices is 4
...

The coefficients of the terms follow an interesting pattern
...
It also enables us to determine the coefficient of any particular term of an
expansion of (a + b)n
...
The coefficients of the expansion of
(a + b)n , for a particular positive integer n, are contained in sequence in the nth row
of this triangle of numbers
...

For example, the triangular numbers occur in Pascal’s triangle along the diagonal shown
in the following diagram
...


{6}

• The binomial theorem
1
1
1

5
6

7

1
1

8

3

10
15

56

1
4

10
20

35

21
28

1

6

4

1
1

2
3

1
1

1

1
5

15
35

70

1
6
7

21
56

1

28

1
8

1

The triangular numbers in Pascal’s triangle
...

The relationship between the expansion of (a + b)n and binomial probabilities is addressed in the module Binomial distribution
...

(1 + x)0 = 1
(1 + x)1 = 1 + x
(1 + x)2 = 1 + 2x + x 2
(1 + x)3 = 1 + 3x + 3x 2 + x 3
(1 + x)4 = 1 + 4x + 6x 2 + 4x 3 + x 4
(1 + x)5 = 1 + 5x + 10x 2 + 10x 3 + 5x 4 + x 5
When the coefficients in the expansions of (1 + x)n are arranged in a table, the result is
known as Pascal’s triangle
...
Some of the patterns of the triangle are more apparent in this form
...


1 Each number is the sum of the two numbers diagonally above it (with the exception
of the 1’s)
...
e
...

3 The sum of the numbers in each row is a power of 2
...
numbers is equal to the sum of the
second, fourth, sixth,
...
(This is not a totally obvious result
...


Example
Expand

1 (1 + x)6

2 (1 − 2x)6
...


2 The expansion of (1 − 2x)6 can be obtained by replacing (−2x) for x in the expansion
of (1 + x)6 :
(1 − 2x)6 = 1 + 6(−2x) + 15(−2x)2 + 20(−2x)3 + 15(−2x)4 + 6(−2x)5 + (−2x)6
= 1 − 12x + 60x 2 − 160x 3 + 240x 4 − 192x 5 + 64x 6
...
With two binomials,
we have
(a + b)(c + d ) = a(c + d ) + b(c + d ) = ac + ad + bc + bd
...
There are 2 × 2 = 22 = 4 terms
...

There are 2 × 2 × 2 = 23 = 8 terms
...


Example
Find the coefficient of x 2 in the expansion of (2x − 1)(3x + 4)(5x − 6)
...
If we do this in all possible ways and add, we will find the required
coefficient
...


Permutations and factorial notation
In how many ways can eight people line up to get into a theme-park ride?
We can draw a box diagram for this situation, with each box indicating the number of
choices for each position
...
The notation for this is 8!
...

Most calculators have a factorial key
...
It is important to note
that n! = n(n − 1)!
...

If there are eight competitors in a race, in how many ways can the first four places be
filled? We can draw a box diagram for this situation, with each box indicating the number
of choices for each position
...


7

6

5

{10}

• The binomial theorem

A permutation is an arrangement of elements chosen from a certain set
...
Some of the permutations of three letters taken from this
set include
abc,

bac,

cab,

ed a
...

The symbol n P r is used to denote the number of permutations of r distinct objects chosen from n objects
...

(n − r )!

Example
List all the permutations of the letters in the word CAT
...


Example
Seven runners are competing in a race
...
The silver medal can then be awarded in 6
ways
...
The total number of ways of
awarding the medals is 7 × 6 × 5 = 210 ways
...
The three letters a, b, c can be
arranged in six ways:
abc,

acb,

bac,

bca,

c ab,

cba
...
The number of permutations of three letters chosen from five letters is 5 P 3 = 5×4×3 = 60
...
We list them here:

{a, b, c},

{a, b, d },

{a, b, e},

{a, c, d },

{a, c, e},

{a, d , e},

{b, c, d },

{b, c, e},

{b, d , e},

{c, d , e}
...

We denote the number of ways of choosing r objects from n objects by

n
, which is read
r

as ‘n choose r ’
...

3
3!
6

In general, we have
n
n
Pr
n!
=
=

...
There are
5
= 5 ways of choosing 1 letter,
1

5
= 10 ways of choosing 2 letters,
2

5
= 15 ways of choosing 3 letters,
3

5
= 10 ways of choosing 4 letters,
4

5
= 1 way of choosing 5 letters
...


Example
Evaluate

1

100
2

2

1000

...

998
2

5
= 1
...

2
2

Example
There are ten people in a basketball squad
...


Solution
10
10 × 9 × 8 × 7 × 6
=
= 252 ways of choosing the starting five
...

2

1 There are

If we expand (a + b)6 , we know the terms will be of the form
c0 a 6 ,

c 1 a 5 b,

c2 a 4 b 2 ,

c3 a 3 b 3 ,

c4 a 2 b 4 ,

c 5 ab 5 ,

c6 b 6 ,

where c i are the coefficients
...

For c 5 , the relevant terms when multiplying out are
abbbbb,
There are

babbbb,

bbabbb,

bbbabb,

bbbbab,

bbbbba
...
Equivalently, there are
1

6
= 6 ways of choosing five b’s from the six brackets
...

5
We can find the values of the other coefficients in the same way
...


A guide for teachers – Years 11 and 12

• {13}

Example
Write out the expansion of (a + b)10
...

Using a similar argument to that given above, we have
c1 =

10
,
1

c2 =

10
,
2

c3 =

10
,
3


...


Exercise 2
Let c i denote the coefficient of the term a 12−i b i in the expansion of (a + b)12
...


The binomial theorem
We are now ready to prove the binomial theorem
...


Theorem (Binomial theorem)
For each positive integer n,
(a + b)n = a n +

n n−1
n n−2 2
n n−r r
n
a
b+
a
b +···+
a
b +···+
ab n−1 + b n
...
If we choose one letter from
each of the factors of
(a + b)(a + b)(a + b) · · · (a + b)
and multiply them all together, we obtain a term of the product
...


• If we choose a from every one of the factors, we get a n
...


• We could choose b from one of the factors and choose a from the remaining
n
n − 1 factors
...
So
1
n n−1
the term with b is
a
b
...
The number of ways of choosing two b’s from n factors is

...

2

• In general, we choose b from r factors and choose a from the remaining n − r
n
factors
...
So the
r
n n−r r
term with b r is
a
b
...
This can be done in
only one way
...

1
2
r
n −1

The binomial theorem can also be stated using summation notation:
(a + b)n =

n
r =0

n n−r r
a
b
...

0
1
2
r
n −1
n

We can now display Pascal’s triangle with the notation of the binomial theorem
...
Numbers of the form

x8

8
8
n
is the coefficient
r

n
are called binomial coefficients
...


Observation 1
Each number in Pascal’s triangle is the sum of the two numbers diagonally above it (with
the exception of the 1’s)
...
In the
notation introduced earlier in this module, this says
5
4
4
=
+

...


{16}

• The binomial theorem

Theorem (Pascal’s identity)
n
n −1
n −1
=
+
,
r
r −1
r

for 0 < r < n
...

r

We will now undertake an alternative proof of Pascal’s identity in a special case, but the
general proof is essentially the same
...

6

...
The
4
subsets with four elements can be separated into two types:
The number of subsets with four elements is

• the four-element sets that contain c
• the four-element sets that do not contain c
...
The number of four-element
3
5
subsets that do not contain c is

...

4
3
4
This gives us a combinatorial proof of Pascal’s identity
...
The question arises as to how many paths there
are from the bottom-left corner to the top-right corner of the grid, if you can only move to
the right or up
...
One path from the bottom-left
corner to the top-right corner is shown in red
...

The numbers on the grid indicate the number of paths to that point
...
You can move up and across the grid using addition to find
the number of paths to any point on the grid
...
There is one path to P
and two paths to X
...

In the same way you can work out all the numbers in the grid, ending with the 20 paths
from D to B
...
We can think of the paths
as strings of letters
...
The
path shown in the diagram is RU RU RU
...
There must be three U ’s and three R’s in each such string
...
So there are
= 20 paths from D to B
...
It takes 2n moves to go from the bottom-left corner of an n × n grid to the topright corner, and n of these moves are U and n are R
...
RU R
...
There are 2n letters in the string and there are
n U ’s and n R’s
...
There are
ways of doing this and so, for an n ×n grid,
n
2n
there are
paths
...

n

Observation 2
Each row of Pascal’s triangle is symmetric
...
This means that the coefficient of x r in the expansion of (1 + x)n is the
same as the coefficient of x n−r
...
In fact, the sum of the entries in
the nth row is 2n
...

0
1
2
r
n −1
n

Substituting x = 1 gives
2n =

n
n
n
n
n
n
+
+
+···+
+···+
+

...
Here is an example to illustrate
the idea
...
How many ways can this choice be made?
Let S = {A, B,C }
...


A guide for teachers – Years 11 and 12

• {19}

There are
3
= 1 set with no elements,
0

3
= 3 sets with one element,
1

3
= 3 sets with two elements,
2

3
= 1 set with three elements
...

0
1
2
3
Now think of each chocolate one at a time:

• chocolate A can be chosen or not (2 ways)
• chocolate B can be chosen or not (2 ways)
• chocolate C can be chosen or not (2 ways)
...
This gives
3
3
3
3
+
+
+
= 23
...


Observation 4
In any row of Pascal’s triangle, the sum of the first, third, fifth,
...
numbers
...

0
1
2
r
n −1
n

Let x = −1
...

0
1
2
3
r
n −1
n

Therefore
n
n
n
n
n
n
+
+
+··· =
+
+
+··· ,
0
2
4
1
3
5
which is the observation
...
Using Observation 3, it
follows that each of these sums is equal to 2n−1
...

0
2
4
6
1
3
5
7

Exercise 4
a Find an identity by substituting x = 2 into the binomial expansion for (1 + x)n
...


Applying the binomial theorem
In this section, we give some examples of applying the binomial theorem
...


Solution
Using the binomial theorem:
(a + b)4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4
...
Then
(2x + 3)4 = (2x)4 + 4(2x)3 × 3 + 6(2x)2 × 32 + 4(2x) × 33 + 34
= 16x 4 + 96x 3 + 216x 2 + 216x + 81
...


Solution
Using the binomial theorem:
(a + b)n =

n
r =0

n n−r r
a
b
...
Then
(1 − x)10 =

10
r =0

10
(−x)r
r

= 1 − 10x + 45x 2 − 120x 3 + 210x 4 − 252x 5 + 210x 6 − 120x 7 + 45x 8 − 10x 9 + x 10
...


The general term in the binomial expansion
The general term in the expansion of (a + b)n is
n n−r r
a
b ,
r

where 0 ≤ r ≤ n
...


Solution
The general term is
5 5−r
x (−3z 2 )r ,
r

where 0 ≤ r ≤ 5
...

Thus the coefficient of x 3 z 4 is

5
(−3)2 = 90
...
How many terms are there in the expansion? Find a
formula for the general term
...


Solution
The general term is
6−r

1
x2

r

=

6 12−2r 1
x
r
x 2r

=

6
x2
r

6 12−4r
x

...

Hence the constant term in the expansion of x 2 +

1
x2

6

is

6
= 20
...


Example
Find the coefficient of x 5 in the expansion of (1 − 2x + 3x 2 )5
...


• The coefficient of x 5 in (1 − 2x)5 is (−2)5 = −32
...

3
1

(−2) = −540
...

Therefore the required coefficient is −32 − 480 − 540 = −1052
...
Let n = 2m, for some positive integer m
...

m
For example, the middle term of (a + b)8 is
8 4 4
a b = 70a 4 b 4
...
Let n = 2m + 1, for some
positive integer m
...

m +1

and

Exercise 8
Find the middle term of (2x − 3y)6
...


When n is odd, there are two greatest coefficients, which are the coefficients of the two
middle terms
n
1
2 (n − 1)

and

n
1
2 (n + 1)


...


{24}

• The binomial theorem

Example
Find the greatest coefficient in the expansion of (1 + 3x)21
...
The
k
k

21
3k+1
...

k +1

To find where the coefficients are increasing, we solve c k+1 > c k , that is,
From above,

c k+1
> 1
...

k +1

Now k is an integer, and hence c k+1 > c k for k = 0, 1, 2,
...

The sequence of coefficients is increasing from c 0 to c 16 and decreasing from c 16 to c 21
...

16

Exercise 9
Find the greatest coefficient of (2x + 3y)15
...
We will need to use Pascal’s identity in the form
n
n
n +1
+
=
,
r −1
r
r

for 0 < r ≤ n
...

1
2
r
n −1

(a + b)n = a n +

We first note that the result is true for n = 1 and n = 2
...
So
(a + b)k = a k +

k k−1
k k−2 2
k k−r r
k
a
b+
a
b +···+
a
b +···+
ab k−1 + b k
...

k −1

From Pascal’s identity, it follows that
(a + b)k+1 = a k+1 +

k +1 k
k + 1 k−r +1 r
k +1
a b +···+
a
b +···+
ab k + b k+1
...
By induction, the result is true for all positive integers n
...


{26}

• The binomial theorem
1
1

1

1

2
3

1
1

6

1
7

1
8

1

3
6

4
5

1

10
15

4

20

56

1

10

35

21
28

1

1
5

15
35

70

1
6
7

21
56

1

28

1
8

1

Sum of the squares for a row of Pascal’s triangle
...

Using the fact that
n
n
=
,
r
n −r
we can write the binomial expansion in two different ways:
(1 + x)n =

n
n
n 2
n r
n
n n
+
x+
x +···+
x +···+
x n−1 +
x ,
0
1
2
r
n −1
n

(1 + x)n =

n
n
n
n n−r
n n−1
n n
+
x+
x2 + · · · +
x
+···+
x
+
x
...


Also, since
(1 + x)n (1 + x)n = (1 + x)2n ,
the coefficient of x n in this expansion is
coefficient of x n we have
2n
n
=
n
0

2

+

n
1

2

+···+

n
r

2n

...


A guide for teachers – Years 11 and 12

• {27}

Using summation notation, this identity is
2

n
2n
n
=
n
r =0 r


...


Exercise 10
The number of forward paths from the bottom left to the top right of an n×n grid is

2n

...


(Hint
...
Choose a point of the grid on the diagonal line
...
Repeat for each point on the diagonal line
...


Exercise 11
Each of the following expansions has three coefficients in arithmetic progression:
(1 + x)7 = 1 + 7x + 21x 2 + 35x 3 + 35x 4 + 21x 5 + 7x 6 + x 7 ,
(1 + x)14 = 1 + 14x + 91x 2 + 364x 3 + 1001x 4 + 2002x 5 + 3003x 6 + 3432x 7 + · · · ,
(1 + x)23 = · · · + 490 314x 8 + 817 190x 9 + 1 144 066x 10 + · · ·
...
In the second
expansion, where n = 14, the progression is 1001, 2002, 3003
...

(Hint
...
)
r −1
r
r +1

{28}

• The binomial theorem

Example
Prove the identity
n

n × 2n−1 =

k×
k=0

n
k

and check in the case when n = 4
...

k

Differentiating both sides of this identity with respect to x gives
n(1 + x)n−1 =

n

k
k=0

n k−1
x

...

k

When n = 4, the left-hand side is 4 × 23 = 32 and the right-hand side is
4+2×

4
4
+3×
+4 = 4+2×6+3×4+4
2
3
= 32
...

k

For example, when n = 6 this gives
6
6
6
6
6
6
+3
+5
=2
+4
+6

...
Then
F (n + 1) =

n +1
n
n −1
+
+
+···
...


Using Pascal’s identity, we obtain
F (n + 1) + F (n) =

n +2
n +1
n
+
+
+ · · · = F (n + 2)
...
is the sequence of
Fibonacci numbers
...


1
1
1

6

1
7

1
1

8

28

3

10
15

56

1
4

10
20

35

21

1

6

4
5

1

2
3

1
1

1

5
15

35
70

1

1
6

1
7

21
56

F8

28

1
8

1 + 7 + 15 + 10 + 1 = 34 = F 8

The Fibonacci numbers in Pascal’s triangle
...
This gives (for n > 0),
n
n
n
n
n
+
+
+
+···+
= 2n
0
1
2
3
n
n
n
n
n
n
+i
−
−i
+···+in
= (1 + i )n
0
1
2
3
n
n
n
n
n
n
−
+
−
+ · · · + (−1)n
= 0n
0
1
2
3
n
n
n
n
n
n
−i
−
+i
+ · · · + (−i )n
= (1 − i )n
...

0
4
8
4
It is easy to show, using the polar form, that
(1 − i )n + (1 + i )n = 2( 2)n cos

nπ

...

+
+
+ · · · = 2n + 2( 2)n cos
4
4
0
4
8
This is a rather amazing and beautiful result
...


The binomial theorem gives
1+

1
n

n

=

n
n 1
n
+
+
0
1 n
2

1
n

2

+· · ·+

n
r

1
n

r

+· · ·+

n
n −1

1
n

n−1

+

n
n

1
n

n


...

0! 1! 2! 3!

Formally, this argument involves some difficult limiting processes
...
It states
that, for |x| < 1,
(1 + x)r =

∞

r (r − 1) · · · (r − k + 1) k
x
k!
k=1

where r is any real number
...

1+x
The expansion for (1 − x)−1 is
1
= 1 + x + x2 + x3 + x4 + · · ·
...
We know that they are only convergent
for |x| < 1
...

The derivative of x n with respect to x is given by
(x + h)n − x n
d (x n )
= lim

...

k

Thus
(x + h)n − x n
= nx n−1
...
For example for 1
...
01)5 ≈ 1 + 5 × 0
...
012 = 1 + 0
...
0001 = 1
...

Of course, with calculators, it is not necessary to go through such a process
...


History
The expansion (a + b)2 = a 2 + 2ab + b 2 appears in Book 2 of Euclid
...

The triangle of numbers that we refer to as Pascal’s triangle was known before Pascal
...

The triangle had been discovered centuries earlier in India and China
...
In China, Pascal’s triangle
is called Yang Hui’s triangle
...
It was also discussed by the Persian poetastronomer-mathematician Omar Khayyam (1048–1131)
...
Several theorems
related to the triangle were known, including the binomial theorem for non-negative integer exponents
...
In Italy, it
is still referred to as Tartaglia’s triangle, named for the Italian algebraist Niccolo Tartaglia
(1500–1577)
...
Pascal’s work on the binomial coefficients led to Newton’s discovery of
the general binomial theorem for fractional and negative powers
...

b There are 25 = 32 terms in the expansion
...

9

Exercise 3
The possible strings of R’s and U ’s have length m+n
...
So the number of ways of choosing the paths is
m +n
m +n
=

...

0
1
2
r
n −1
n

When x = 2, we have
3n = (1 + 2)n
=

n
n
n 2
n r
n
n n
+
2+
2 +···+
2 +···+
2n−1 +
2
0
1
2
r
n −1
n

= 1+2

n
n
n
n
+4
+ · · · + 2r
+ · · · + 2n−1
+ 2n
...

1
2
r
n −1

Exercise 5
a (2x − 3y)4 = 16x 4 − 96x 3 y + 216x 2 y 2 − 216x y 3 + 81y 4
b

x−

2
x

4

= 24 +

16 32
−
− 8x 2 + x 4
x4 x2

Exercise 6
There are eight terms in the expansion
...

r

Exercise 7
The constant term is 15
...


Exercise 9
c 9 = 26 × 39

15
= 6 304 858 560
9

Exercise 10
In the following diagram, there are

3+2
3

paths from Y to X, and

3+2
2

paths from X to W
...


Every path from Y to W meets the diagonal, and so the total number of paths from Y to W
can be found by summing the number of such paths through each point on the diagonal
...


A guide for teachers – Years 11 and 12

• {35}

W

X
Y

Exercise 11
Assume that the coefficients of three consecutive powers of (1 + x)n are in arithmetic
progression
...

r +1

Then
n
n
n
n
−
=
−
r
r −1
r +1
r
n!
n!
n!
n!
−
=
−
(n − r )!r ! (n − r + 1)!(r − 1)! (n − r − 1)!(r + 1)! (n − r )!r !
n!
1
1
n!
1
1
−
=
−
(n − r )!(r − 1)! r n − r + 1
(n − r − 1)!r ! r + 1 n − r
n − 2r + 1 n − 2r − 1
=
n −r +1
r +1
(n − 2r )2 = n + 2
...


0

1

2

3

4

5

6

7

8

9

10

11

12



---