Mathematics B30
Module 4
Lesson 14
Mathematics B30
The Binomial Expansion
69
Lesson 14
Mathematics B30
70
Lesson 14
The Binomial Expansion
Introduction
By direct multiplication it is not too difficult or time consuming to expand binomials with
2
3
4
low powers such as  x  y  ,  x  y  , or  x  y  .
It would be too tedious to expand higher powers of binomials such as  x  y 
direct multiplication.
10
or  x  y 
20
by
Fortunately, the expansion of the first few low powers of binomials provide a simple pattern
which can be followed for expanding higher powers.
In this lesson a method of expanding binomials with large powers is studied. You will also
see applications of this to probability problems of the special kind where each trial has only
two possible outcomes.
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Lesson 14
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Lesson 14
Objectives
After completing this lesson you will be able to:
•
determine the coefficients of terms in a binomial expansion using
the Binomial Theorem.
•
expand binomials of the form (a +b)n, for any positive integer n,
using the Binomial Theorem.
•
solve word problems involving the expansion of binomials.
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Lesson 14
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Lesson 14
14.1
Pascal’s Triangle
By direct multiplication it is not too difficult or time consuming to expand binomials with
2
3
4
low powers such as  x  y  ,  x  y  , or  x  y  . It would be too tedious to expand higher
powers of binomials such as  x  y  or  x  y  by direct multiplication. Fortunately, the
expansion of the first few low powers of binomials provides a simple pattern which can be
followed in expanding high powers.
10
20
The pattern is established in the following examples. Each expansion is obtained by
multiplication as shown.
 x  y 2  x  y x  y   x 2  2 xy  y 2 ,
 x  y 3   x  y 2  x  y   x 2  2 xy  y 2  x  y 
 x 3  3 x 2 y  3 xy 2  y 3 , etc.
 x  y 0
 x  y 1
 x  y 2
 x  y 3
 x  y 4
1
 1x  1 y
 1 x 2  2 xy  1 y 2
 1 x 3  3 x 2 y  3 xy 2  1 y 3
 1 x 4  4 x 3 y  6 x 2 y  4 xy 3  1 y 4
The coefficients of the terms can be arranged into what is called Pascal’s Triangle where any
number in the triangle is found by adding the two numbers above on either side. Each row
begins and ends with the number 1.
 x  y 0
 x  y 1
 x  y 2
 x  y 3
 x  y 4
 x  y 5
1
row 0
1
row 1
1
row 2
row 4
row 5
4
1
1
2
3
1
row 3
1
5
1
3
6
10
4
10
Coefficients
1
1
5
1
(and the Triangle continues)
According to the pattern the sixth row, or the coefficients of the terms of  x  y  , should be
1 6 15 20 15 6 1
6
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Lesson 14
By observing the previous expansion, other patterns are found.
Here is the pattern for expansion of  x  y  .
n
• If the power of the binomial is n, the expansion has n  1 terms.
• The first term is x n and the last is y n .
• Going from left to right the powers of x decrease by one from n to 0 and the powers of
y increase by one from 0 to n.
• In each term the sum of the exponents of x and y is n.
Example 1
By using the patterns write the expansion of  x  y  .
6
Solution:
Determine the coefficients.
(Pascal's Triangle)
1 6 15 20 15 6 1
Determine the number of terms
for the expansion.
n  1, n  6
 7 terms
Determine the terms.
x6 y0
x 5 y1
x4 y2
x3 y3
x2 y4
x1 y5
x0 y6
Combine the terms and the coefficients
with plus signs.
 x  y 6
 1 x 6  6 x 5 y  15 x 4 y 2  20 x 3 y 3  15 x 2 y 4  6 xy 5  1 y 6 .
Example 2
Use Pascal’s Triangle to write the expansion of  x  y  .
7
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Lesson 14
Solution:
Determine the coefficients.
1 7 21 35 35 21 7 1
Determine the number of
terms required.
n  1, where n  7
 8 terms
Determine the terms.
x7 y0
x 6 y1
x5 y2
x 4 y3
x3 y4
x 2 y5
x1 y6
x 0 y7
Introduce the coefficients and the
plus signs.
( x  y)7  x 7  7 x 6 y  21 x 5 y 2  35 x 4 y 3  35 x 3 y 4  21 x 2 y 5  7 xy 6  y 7
Note carefully how the minus sign is treated in the next example.
Example 3
Expand  x  y  .
5
Solution:
( x  y) 5 may be written as  x   y  .
5
This expansion is identical to that of  x  y  , except that y is replaced by  y  .
5
x  y 5  x   y 5
 x 5  5 x 4  y   10 x 3  y   10 x 2  y   5 x  y    y 
2
3
4
5
 x 5  5 x 4 y  10 x 3 y 2  10 x 2 y 3  5 xy 4  y 5
The signs alternate because  y   y 2 ,  y    y 3 ,  y   y 4 ,  y    y 5 .
2
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77
4
5
Lesson 14
The binomial  x  y  may have further variations. The x or the y or both may be replaced
n


by more complicated expressions such as in 2 a  b 2 . This is of the form  x  y  with
x  2 a and y   b 2 . The same pattern for expansion applies but simplification of terms is
required.
4
4
Example 4
4
 2x 1 
Simplify 
  .
 y x
Solution:
Determine the coefficients.
1 4 6 4 1.
The expansion before simplification is
4
3
2
2
3
4
 2x 
 2x   1   2x   1 
 2 x  1 
1
1   4      6      4     1  .
 x
 y 
 y   x  y   x
 y  x 
4
x
x2
1
1
1
16    32 3  24 2  8 2  4 .
y
y
yx
x
 y
Simplify.
Exercise 14.1
1.
Using the pattern for forming successive rows of Pascal’s Triangle, write out the 7th to
the 10th rows of the triangle.
2.
Fill in the missing numbers in this portion of Pascal’s Triangle.
5
5
•
•
•
4
6
2
•
2
2
0 4
9
5 •
•
1
2
8
7 •
•
3.
•
Add the numbers in each of the rows of Pascal’s Triangle. What pattern appears in
the sums?
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Lesson 14
4.
Explain why 1 3 7 3 1 could not be a row in Pascal’s Triangle.
5.
Without specifying the coefficients, list the terms in x and y used in the expansion of
a.
b.
6.
 x  y 7
2 x  y 7
Expand and simplify each of the following. Note that one variable is replaced by a
constant in some binomials.
a.
 x  y 7
b.
2  t 
c.
x  1
d.
1  y 5
5
4
e.
 x  2 5
f.
y

2x  
2

g.
1 

x  2 
x 

4
4
7.
Write the complete 4th term in the expansion of a  b  .
8.
Without regard to the coefficient write the 4th term in the expansion of
8
a.
b.
c.
a  b 16
 x  y 20
 x  y 13
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Lesson 14
14.2
The Binomial Theorem
Pascal’s Triangle does make it an easy matter to find the coefficients in the expansion of
 x  y n for relatively small integers n. In order to be able to expand  x  y n for large n in
an efficient manner, an improvement over Pascal’s Triangle must be found.
To this end we will look for further patterns in the coefficients. The pattern will be seen
when we write the coefficients in another way.
•
Coefficients in expansion of  x  y  .
1
5
10
10
1
5
1
5× 4
1× 2
5× 4× 3
1× 2× 3
•
Coefficients in expansion of  x  y  .
1
5
15
20
15
6
1
6
1
6× 5
1× 2
6× 5× 4
1× 2× 3
6× 5× 4× 3
1× 2× 3× 4
6× 5× 4× 3× 2
1× 2× 3× 4× 5
5
5
1
5× 4× 3× 2
1× 2× 3× 4
5× 4× 3× 2× 1
1× 2× 3× 4× 5
6
1
6× 5× 4× 3× 2× 1
1× 2× 3× 4× 5× 6
In each case the numerator and denominator have one fewer factors than the
number of the term.
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Lesson 14
The pattern for coefficients is as follows.
• The first and last terms have coefficient 1.
n
• The rth term in the expansion of  x  y  is made up of r  1 factors in the
numerator and r  1 factors in the denominator.
n n  1 n  2     n  r  2 
1  2  3    r  1 
The numerator factors decrease - starting from n - and the denominator factors increase,
starting from 1.
For example, the coefficient of the 4th term in the expansion of  x  y 
20
20  19  18
 1 140 .
123
*Notice that this is also
The coefficient of the 7th term in the expansion of  x  y 
16
16  15  14  13  12  11
 8 008 .
1234 56
20
is
C3 .
is
*Notice that this is
16
C6 .
In general, the coefficient of the rth term of  x  y  is n C r 1 .
n
Example 1
Write the first 6 terms of the expansion of  x  y  .
21
Solution:
The powers of x decrease from 21 and the powers of  y  increase from 0.
 x  y 21   x   y 21
0
1
2
3
 21 C 0 x 21  y   21 C1 x 20  y   21 C 2 x 19  y   21 C 3 x 18  y  
4
5
17
16
21 C 4 x  y   21 C 5 x  y     
 x 21  21 x 20 y  210 x 19 y 2  1 330 x 18 y 3  5 985 x 17 y 4  20 349 x 16 y 5    
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Lesson 14
Example 2
Write the first 4 terms in the expansion of (2 x  3 y)7 .
Solution:
7
7 6
7 6 5
(2 x )7  (2 x ) 6 ( 3 y)1 
(2 x ) 5 ( 3 y) 2 
(2 x ) 4 ( 3 y) 3
1
12
123
7 7
6
6
5
2 5 2
 2 x  7(2 )( 3) x y  21(2 )( 3) x y  35 (2 4 )( 3) 3 x 4 y 3
 128 x 7  1344 x 6 y  6048 x 5 y 2  15120 x 4 y 3 .
The full Binomial Theorem is now stated and summarized.
Binomial Theorem: If n is a positive integer, then
 x  y n
 xn 
n n 1 1 n n  1  n 2 2 n n  1 n  2  n 3 3
x y 
x y 
x y    yn
1
12
123
This may also be written with the coefficients in
combinatorial form, n Cr .
The rth term of the expansion of  x  y  is
n
n n  1 n  2     n  r  2  n r 1  r 1
x
y  n C r 1 x n r 1  y r 1
1  2  3    r  1 
Study the pattern in the expansion very carefully. The following is an important
summary of the pattern.
•
•
•
The power of y is one less than the number of the term.
The power of x is such that the sum of the exponents of the two powers is n.
The numerator and denominator of the coefficient of the rth term each have
r  1 factors; the numerator, a product of consecutive integers starting from n
and going down, the denominator a product of consecutive integers starting
from 1 and going up.
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Lesson 14
Example 3
Find the middle term or terms in the expansion of
a.
b.
1  2 y 13
 x  2 14
Solution:
a.
The expansion of 1  2 y  has 14 terms. The middle of this expansion must be both
the 7th and 8th terms with 6 terms on either side.
13
The 7th term is
13  12  11  10  9  8 7
1  2 y 6  1 716  2 6 y 6
1234 56
 109 824 y 6
The 8th term is
13  12  11  10  9  8  7 6
1  2 y 7  1 716  2 7 y 7
1  2  3  4  5  6 7
 219 648 y 7
b.
The expansion of  x  2  has 15 terms. The middle of this expansion is the 8th term
only, with 7 terms on either side.
14
The 8th term is
14  13  12  11  10  9  8 7
7
x  2   3 432  128  x 7
1  2  3  4  5  6 7
 439 296 x 7
Example 4
Which term of the expansion of a  b 
n C r form?
54
contains a 17 b 37 and what is its coefficient in
Solution:
Since the power of b is 37 this must be the 38th term. Its coefficient is
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54
C 37 .
Lesson 14
Example 5
What is the coefficient of a 4 b7 in the expansion of 2 a  b  ?
11
Solution:
Since the power of b is 7, a 4 b 7 must be part of the 8th term.
The 8th term is
11  10  9  8  7  6  5
2 a 4  b7  330 2 4 a 4 b7
1  2  3  4  5  6 7
 5 280 a 4 b 7
 
The coefficient of a 4 b7 is  5 280 .
Exercise 14.2
1.
Evaluate the coefficients of the indicated terms.
a.
b.
c.
d.
e.
2.
5th term of  x  y 
7
3rd term of  x  y 
7
4th term of  x  y 
25
middle term of  x  y 
12
2 middle terms of  x  y 
13
Write and simplify the first 3 terms of the expansion of
a.
b.
c.
d.
e.
 x  y 50
1  y 
2 20
 x  3 y 21
1 

3  a 
3 

1  2 y 13
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Lesson 14
3.
Consider the binomial  x  y  ,
31
a.
b.
4.
What is the degree of each term in the expansion?
How many terms are there in the expansion?
In the binomial expansion of  x  y  a term containing x 7 y 15 occurs.
n
a.
b.
c.
What is the value of n?
What is the coefficient of x 7 y 15 ?
How many terms are in the expansion?
5.
In the expansion of  x  y 
x 10 y 7 ?
which other term has the same coefficient as the 8th term
6.
In the expansion of  x  y 
there is a term containing x k y 14 .
17
26
a.
b.
7.
Expand and simplify each of the following.
a.
b.
c.
8.
What is the value of k?
Find the coefficients of x k y 14 and x 14 y k .
2 

c  2 
c 

5
 3 1 
x  2 
x 

2 a  3 a 1

4

3
Find the middle term(s) of the expansion of each binomial.
a.
b.
c.
2 a  3 b 17
a

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1
a
2 i


20
7
85
Lesson 14
9.
Determine the coefficients of the terms that contain these variables.
a.
2 x  y 10 ,
x7 y3
7
b.
c.
10.
y

 3x   ,
3

2 a  3 b 9 ,
x2 y5
a 5 b4
Simplify.
a.
 x  h 5  x 5
h
2x  h   2 x 4
h
4
b.
14.3
The Binomial Expansion and Probability of
Events with only Two Possible Outcomes
The problems in this section deal with events that have two possible outcomes and each
outcome has an equally likely chance of occurring. Usually several trials are conducted in
which one or the other of the outcomes will occur.
For example, a coin is tossed (this is one trial) and the two events are that of a Head landing
up or a Tail landing up. If the coin is fair, each event has an equally likely chance of
occurring; i.e., the probability of each event is 0.5.
Another example involves a young couple planning to have a family of four children. Having
one child is called a trial and the two events are having a boy or having a girl. Each of these
events may be considered equally likely. What is the probability of having exactly 2 boys
and 2 girls? What is the probability of having exactly four girls? These questions can be
answered using one of several methods. First we answer the question by making use of the
tree diagram and then by using what is known about expanding binomials.
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Lesson 14
Trial
1
2
3
B
B
G
B
B
G
G
B
B
G
G
B
G
G
4
B
G
B
G
B
G
B
G
B
G
B
G
B
G
B
G
The sample space consists of 2  2  2  2  16 different kinds of families. By a count, there
are 16 different families of 2 boys and 2 girls.
P(2 boys and 2 girls) =
Similarly, P(4 girls) =
6
3
 .
16 8
1
16
Now proceed to relate finding this probability to a binomial expansion.
Let the probability of having a boy be b and the probability of having a girl be g. By the
First Principle of Counting the probability of any one of the families (branches) occuring is
the product of the probabilities of the events on the branch.
The probability of 4 boys is b  b  b  b  b 4 . The probability of BGBG is bgbg  b 2 g 2 etc.
Refer to the number of branches in the tree above.
There is only one way of having exactly 4 boys. Therefore, P 4 boys   1  b 4 or 4 C 0 b 4 .
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Lesson 14
There are 4 
4
4!
 4 C 1 ways of having 3 boys and 1 girl. Therefore, P(3 boys, 1 girl) =
3!1!
C1 b 3 g .
There are 6 ways of having exactly 2 boys and 2 girls in a family.
Therefore, P 2 boys, 2 girls   4 C 2 b 2 g 2 .
Similarly, P 1 boy, 3 girls
  4 C3 bg 3 , and P 4 girls   4 C 4 g 4 .
If the probabilities of all possible families are added up you get
4
C0 b 4  4 C1 b 3 g  4 C 2 b 2 g 2  4 C3 bg 3  4 C 4 g 4 , which is nothing more than the expansion of the
binomial b  g  .
4
Assume that b  g 
1
. The probabilities of 2 boys and 2 girls in a family is
2
2
4!  1   1 
   
4 C2 b g 
2!2!  2   2 
 1  1 
 6   
 4  4 
6

16
3

8
2
2
2
Example 1
What is the probability of having at least 2 girls in a family of 4 children?
Solution:
Let b = probability of a boy and g = probability of a girl.
1
1
Assume that b  and g  .
2
2
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Lesson 14
The probability of having at least 2 girls is the sum of the probabilities of having 2, 3, or 4
girls. These are three mutually exclusive events, thus the Second Principal of Counting
applies and the probabilities are added.
4
4
4
11
1
1
1
P (2 girls )  P (3 girls )  P (4 girls )  4 C 2 b g  4 C 3 bg  4 C 4 g  6     4     1    
16
2
2
2
2
2
3
4
In the previous example each occurrence had only two possible outcomes, i.e., either a boy
was born or a girl was born.
The following examples deal with tossing a coin. Here also, there are only two possible
outcomes; it must land head up or tail up. If the coin is fair, there is an equal chance of head
landing up or tail landing up. The binomial expansion is used to determine probabilities
related to tossing a coin.
Example 2
A fair coin is tossed 6 times. Calculate the following probabilities:
a.
b.
c.
3 tails and 3 heads.
5 heads and 1 tail
4 tosses of 1 kind and 2 of the other kind.
Solution:
a.
Let T be the event that tails lands up and H the event that heads lands up and the
1
probability of each is t  h  .
2
6
•
In the expansion t  h  , the coefficient of t 3 h 3 is 6 C3 (the 4th term.)
6
C3 
6 5 4 
 20
3 2 1 
P(3 tails, 3 heads)  6 C 3 t 3 h 3
1
 20  
2
20

64
5

16
Mathematics B30
6
89
Lesson 14
b.
The coefficient of t 1 h 5 in the expansion is 6 C5 (6th term).
6
6
3
1

P(1 tail, 5 heads)  6 C 5 t h  6   
64 32
2
5
Alternate solution:
5
1 1
P (5 heads , 1 tail )  6 C1    
2 2
3

32
c.
1
This can be stated as 4 tails and 2 heads or 2 tails and 4 heads.
C 4 15

64
64
P(4 tails, 2 heads) 
6
P(2 tails, 4 heads) 
6
C 2 15

64
64
Since there are mutually exclusive events, their probabilities must be added.
P(4 of 1 kind, 2 of another) 
15 15 30 15



64 64 64 32
There are numerous other examples of events in which there are exactly 2 possible
outcomes.
Example 3
If a skier has a 40% chance of injury on any one jump what is the probability of not
being injured after three jumps?
Solution:
Let q(probability for injury) 
2
3
and p(probability of no injury)  .
5
5
Since there are only two possible outcomes, this is a binomial situation in which n  3 .
q  p 3
Mathematics B30
 q 3  3 q 2 p  3 qp 2  p 3
90
Lesson 14
3
27
3
 0 .216 .
The probability of no injury after 3 jumps is p    
125
5
3
Mathematics B30
91
Lesson 14
Exercise 14.3
1.
Suppose the likelihood of having a boy is the same as that of having a girl for a family
of five children.
a.
b.
c.
In how many ways is it possible to have three girls and two boys?
What is the probability of having a family of 3 girls and 2 boys?
What is the probability of having 4 girls and 1 boy?
2.
In a family of 7 children what is the probability of having at least 5 girls?
3.
A coin is tossed 3 times. Calculate the probability of the following events.
a.
b.
c.
3 heads
at least 2 heads
not more than 2 heads
4.
Illustrate how many different ways a coin tossed seven times could land using the
7
expansion of  H  T  .
5.
In a true and false test of twelve questions how many different ways can 7 true and 5
false questions be arranged? How many different test solutions are possible?
6.
A certain coin which is not fair has a probability of
3
2
of landing heads up and
of
5
5
landing tails up. In 4 tosses of that coin, what is the probability of having exactly 2
heads land up?
Mathematics B30
92
Lesson 14
Mathematics B30
93
Lesson 14
Self Evaluation
1.
2.
Expand and simplify
a.
 2 a  b 
b.
3x
2 4


6
Find, but do not simplify, the middle term(s) of
a.
b.

3  2 iy
2 i  x 12

17
3.
Find the coefficient of x 5 y 7 in the expansion of 2 x  y  .
4.
What is the probability that a fair coin tossed eight times will produce
exactly 5 heads up?
12
Mathematics B30
94
Lesson 14
Mathematics B30
95
Lesson 14
Summary – Lesson 14
•
Create a summary of this lesson to assist you come examination time.
•
Each summary is to be sent in with the assignment to be evaluated.
•
Items to include in a summary:
• definitions
• formulas
• calculator “shortcuts”
Mathematics B30
96
Lesson 14
Mathematics B30
97
Lesson 14
Answers to Exercises
Exercise 14.1
1.
210
2.
55
220
165
715
330
495
2002
1287
792
3003
462
1716
924
462
3.
The nth row has sum 2 n .
4.
The numbers add up to 15 which is not a power of 2.
5.
a.
b.
x 7 , x 6 y, x 5 y 2 , x 4 y 3 , x 3 y 4 , x 2 y 5 , xy 6 , y 7
128 x 7 , 64 x 6 y, 32 x 5 y 2 , 16 x 4 y 3 , 8 x 3 y 4 , 4 x 2 y 5 , 2 xy 6 , y 7
6.
a.
b.
c.
d.
x 7  7 x 6 y  21 x 5 y 2  35 x 4 y 3  35 x 3 y 4  21 x 2 y 5  7 xy 6  y 7
e.
f.
g.
Mathematics B30
32  80 t  80 t 2  40 t 3  10 t 4  t 5
x4  4 x3  6 x2  4 x 1
1  5 y  10 y 2  10 y 3  5 y 4  y 5
x 5  10 x 4  40 x 3  80 x 2  80 x  32
1 4
16 x 4  16 x 3 y  6 x 2 y 2  xy 3 
y
16
6
4
1
x4  4 x  2  5  8
x
x
x
98
Lesson 14
7.
56 a 5 b 3
8.
a.
b.
a 13 b 3
x 17 y 3
c.
x 10  y 
a.
b.
c.
d.
e.
C4  35
7 C 2  21
25 C 3  2 300
12 C 6  924
7th and 8th terms
a.
50
C 0 x 50  50 C1 x 49 y  50 C 2 x 48 y 2  x 50  50 x 49 y  1225 x 48 y 2
b.
20
C 0 1 20  20 C1 119 y 2  20 C 2 118 ( y 2 ) 2  1  20 y 2  190 y 4
c.
21
C 0 x 21  21 C 1 x 20 3 y   20 C 2 x 19 3 y   x 21  63 x 20 y  1 890 x 19 y 2
3
Exercise 14.2
1.
2.
7
13
C6  1 716
13
C7  1 716
 
2
2
 
25
C0 3
e.
13
C 0 1 13  13 C1 1 12 2 y   13 C 2 1 11 2 y   1  26 y  312 y 2
3.
a.
b.
31
32
4.
a.
b.
c.
22
22 C15
23
Mathematics B30
 
a
a
 25 C 1 3    25 C 2 3 23    3 25  25 3 23 a  300 3 21 a 2
3
3
d.
25
24
2
99
Lesson 14
5.
C7 is the coefficient of the eighth term.
coefficient of the 11th term.
6.
a.
b.
12
k 14
26 C14 is the coefficient of x y
14 12
y
26 C12 is the coefficient of x
7.
a.
40 80 80 32



c c 4 c 7 c10
4
1
x 12  4 x 7  6 x 2  3  8
x
x
54
27
8 a 3  36 a 

a a3
17
b.
c.
8.
a.
The 9th and 10th terms are 17 C 8 2 a   3 b  and
or 24 310 2 9 3 8 a 9 b 8 and  24 310 2 8 3 9 a 8 b 9
b.
The 11th term is
c.
The 4th and 5th terms are 7 C 3
a.
b.
c.
10.
a.
b.
Mathematics B30
C7 17 C10 which is the
c 5  10 c 2 
9
  
7
9.
17
C4
 2   i 
3
4
8
  
20
 
C10 a 1
10
C 9 2 a   3 b 
8
17
9
a 10  20 C10  184 756
 2   i 
4
3
 140 i and
 70 2
C 3 2 x  y 3  15 360 x 7 y 3
7
This is the 6th term. The coefficient is .
9
th
This is the 5 term. The coefficient is 326 592.
This is the 4th term.
7
10
x 5  5 x 4 h  10 x 3 h 2  10 x 2 h 3  5 xh 4  h 5  x 5
h
4
3
2 2
 5 x  10 x h  10 x h  5 xh 3  h 4


2 x 4  4 x 3 h  6 x 2 h 2  4 xh 3  h 4  2 x 4
h
3
2
 8 x  12 x h  8 xh 2  2 h 3
100
Lesson 14
Exercise 14.3
a.
B 2 G 3 is the 4th term is the expansion of B  G 
whose coefficient is 5 C3  10
1.
5
2
3
10
5
1 1
10 B G  10     

32 16
2 2
BG 4 is in the 5th term.
b.
2
c.
3
4
5 C 4 BG
4
5
 1  1 
 5    
32
 2  2 
2.
In the expansion of B  G  the terms 7 C 5 B 2 G 5 , 7 C 6 BG 6 , and
7
give the required probabilities. These are mutually
7 C7 G
exclusive events and the probabilities are added to give
21
7
1
29



.
128 128 128 128
3.
a.
7
1
8
1
2
7
8
b.
c.
4.
7
C0  7 C1  7 C2  7 C3  7 C4  7 C5  7 C6  7 C7
These are the coefficients of the expansion of 1  1  .
7
Therefore, the number of ways is 1  1   2 7  128 .
7
5.
C  792
2 different test solutions are possible.
12 5
12
2
6.
Mathematics B30
2
216
2 3
 6    
4 C2 t h
625
5 5
2
2
101
Lesson 14
Answers to Self Evaluation
1.
a.
b.
16 a 4  32 a 3 b 2  24 a 2 b 4  8 ab 6  b 8
27  54 3 x  135 x 2  60 3 x 3  45 x 4  6 3 x 5  x 6
2.
a.
The 9th and 10th terms are
17
b.
C8
 3   2iy 
9
8
The 7th term is
12
and
 25 344
4.
7
1 1
5
3
 56     
8 C3 H T
32
2 2
Mathematics B30
 3   2iy 
8
9
C 6 2 i   x 
3.
5
C9
17
6
6
3
102
Lesson 14
Mathematics B30
103
Lesson 14