Integrated Mathematics Sec 4 (2012)
Term 1:
Topic : Unit 3 Binomial Theorem
Name : __________________ (
)
Class : ___________
Date : ________
3.1 Pascal’s Triangle
Pascal’s Triangle looks like this
1
1
1
1
1
1
2
1
3
3
4
1
6
4
1
When we expand expression such as (x + y)4 we find it follows the same pattern.
4
2
2
Example 1.
(x + y) = x + 2xy + y
Notice the pattern of the coefficients (1,2,1)
Example 2.
3
3
2
2
(x + y) = x + 3x y + 3xy + y Here the coefficients are (1,3,3,1)
The key is to notice the 3 patterns:1. The coefficients follow Pascal’s triangle sequence
2. The powers of one decrease while the other increase
3. The total power in each expression adds up to the original power in the questions
5
Example 3.
5
4
3 2
3 2
4
5
(x + y) = x + 5x y + 10x y + 10x y + 5xy + y
So now we know this:
0
Line 1
1
Line 2
(a + b) = 1
(a + b) = 1a + 1b
2
2
2
3
3
2
2
4
4
3
2 2
(a + b) = 1a + 2ab + 1b
Line 3
3
(a + b) = 1a + 3a b + 3ab + 1b
Line 4
3
4
(a + b) = 1a + 4a b + 6a b + 4ab + 1b
Example 4.
Line 5
Expand (2x + 3y)3 using Pascal’s triangle
4th line is 1,3,3,1
3
0
2
1
1
2
0
3
= 1(2x) (3y) + 3(2x) (3y) + 3(2x) (3y) + 1(2x) (3y)
= 8x + (3  4x  3y) + (3  2x  9y ) + (1  27y )
3
3
2
2
2
2
3
= 8x + 36x y + 54xy + 27y
3
Example 5.
Fully expand (1 + 3x)(1 + 2x)3
3
3
0
2
1
1
2
0
3
(1 + 2x) = 1(1) (2x) + 3(1) (2x) + 3(1) (2x) + 1(1) (2x)
2
3
2
3
= 1 + 6x + 12x + 8x
3
2
3
(1 + 3x)(1 + 2x) = (1 + 3x)(1 + 6x + 12x + 8x )
2
3
= 1 + 6x + 12x + 8x + 3x + 18x + 36x + 24x
2
3
= 1 + 9x + 30x + 44x + 24x
4
4
3.2 Factorial Notation
n! = n  (n – 1)  (n – 2) ......
A factorial notation looks like this
Example 1.
8! = 8  7  6  5  4  3  2  1
To find the number of ways of choosing r items from a group of n items is written as
n
 n 
Cr or  
 r
This is calculated by
Example 2.
Find
5
n!
r ! n  r !
C3
=
5!
2!  3!
=
54321
21324
=
120
26
= 10
Example 3.
Find  7 
 
4
=
7!
3!  4!
=
7654321
3214321
= 35
Example 4.
4 people need to sit down but 2 want to sit together. How many different
combinations are there?
n
Using
4
Cr = C2
=
4!
2!2!
=
4321
2121
=6
3.3 The Connection between Combination Notation and Pascal’s Triangle
Investigate these values :
4
4
C0 = 1
C1 = 4
4
4
C2 = 6
4
C3 = 4
C4 = 1
These values are the same as the Index 4 line
 6th line would be
6 6 6 6 6 6 6
             
0 1 2 3 4 5 6
3.4 Using Factorial Notation to work out the Coefficients in the Binomial Expansion
The Binomial Expansion is:n
(a + b) = (a + b)(a + b)(a + b)........(a + b)
n
n
0
n
n–1 1
= ( C0)a b + ( C1)a
or
Example 1.
n
n–2 2
b + ( C2)a
n
n–n n
b + ...........( Cn)a
b
n n 0
 n  n–1 1
 n  n–2 2
 n  n–n n
b +  a
b ...........   a
b
 a b +  a
0
1
2
n
Use the binomial theorem to find the expansion of (3  2 x)5 .
5 5
5 4
5 3
5 2
0
1
2
3
=   3 ( – 2x) +   3 ( – 2x) +   3 ( – 2x) +   3 ( – 2x)
0
1
2
3
5 1
5 0
4
5
+   3 ( – 2x) +   3 ( – 2x)
4
5
2
3
4
= 243 – 810x + 1080x – 720x + 240x – 32x
5
Example 2.
x

a) Write down the first 4 terms of the expansion 1  
 10 
6 
6 
x 
=   (1)  –

10 
0 

0
6 
5 
x 
+   (1)  –

10 
1 

1
6
6 
4 
x 
+   (1)  –

10 
2 

2
6 
3 
x 
+   (1)  –

10 
3 

2
3



x 
x 
x 


= 1 + 6 
+
15
+
20
–





10 
100 
1000 



2
3
6x 15x
20x
= 1 –
+
–
10
100
1000
2
= 1 – 0·6x + 0·15x – 0·02x
3
b) By substituting an appropriate value of x, find an approximate value to (0.99)6, giving answer to
5 d.p.
This means we want

x 
1–
 = 0·99
10


0·01 =
x
10
0·1 = x
2
3
substitute x = 0.1 into (1 – 0·6x + 0·15x – 0·02x )
= 1 – (0·6  0·1) + (0·15  0·1 ) – (0·02  0·1 )
2
3
= 1 – 0·06 + 0·0015 – 0·00002
= 0·94148
6
Using a calculator 0·99 = 0·941480149
so approximation is accurate to 5dp as this how far the two answers are the same
3.5 Expanding Polynomials Using the Binomial Expansion
You need to remember these facts…
n n
 n  n–1 1
 n  n–2 2
n
(1 + x) =   1 +   1
x +   1
x .........etc
0
1
2
2
or
= 1 + nx +
3
n(n – 1)x
n(n – 1)(n – 2)x
+
........etc
2!
3!
3
(1 + 2x) = 1 + (5  2x) +
5
Example 1.
2
5  4  (2x)
+
3
5  4  3  (2x)
2!
+
3!
2
5
5  4  3  2  (2x)
5  4  3  2  1  (2x)
+
4!
5!
2
3
4
5
= 1 + 10x + (10  4x ) + (10  8x ) + (5  16x ) + (1  32x )
2
3
4
= 1 + 10x + 40x + 80x + 80x + 32x
5
3.6 Taking Out the Common Factors
The Binomial Expansion only works for (1 + x)n so sometimes you need to take out a factor first.
 
3
x  
(2 + x) =  2  1 +  
2 
 
3 
x 
=2 1+ 
2

3
3
Expand as normal
Remember to
multiply by this at
the end
3.7 Finding a particular term in Binomial Expansion
The (r+1) th term of the expansion of (a  b)n is:
 n
Tr 1    anr br
r
Example 1: Find the 6th term of (4x - 5)10.
T6  T51
10 
T6    (4 x)5 ( 5)5
5
T6  806400000 x 5
Homework : Text book Exercises
Ex. 6.1 : Q1,Q2,Q4,Q5,Q7,Q9
Ex. 6.2 : Q3, Q4, Q5,Q6,Q7,Q10