Binomial Expansion
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Transcript Binomial Expansion
Want to know some BRAIN BLOWING MATHS?
1
5
10
10
5
1
The coefficients in the binomial expansion of
(1 + x) 5.
The coefficient of x 6 in the expansion of (1 + x) 49 is
49 C ,
6
That’s the number of ways of winning the jackpot on
the National Lottery.
The number of ways of winning the jackpot on
the National Lottery is 13 983 816
49
C6 13,983,816
13 983 816 two pence pieces laid end to end
would stretch 220 miles –
from London to Paris.
………………
…………………….
13 983 816 seconds is 161 days –
from 11th April until 19th September.
The Binomial Expansion
For Positive Integers
Lets start at the very beginning!
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www.mathsathawthorn.pbwiki.com
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2 16
4
1. What is the value of two to the power of four ?
2. What is the cube of 2x ?
NOT
2x
(2 x)3 8x3
3
(5 y ) 2 25 y 2
3. What is the square of 5y ?
4. Simplify
3 8 x 3 25 y 2 600 x 3 y 2
3 (2 x )3 (5 y ) 2
5. What is the cube of
3
8
2
x3
x
2
x
6. What is any number to the power of zero ?
7. What is ( a + b ) to the power of zero ?
8. Expand and simplify
( a b)
9. Expand and simplify
( a b) 2
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a0 1
( a b) 0 1
ab
That’s all we can do
4
( a b) ( a b)( a b)
2
ANOTHER WAY…..
ONE WAY…..
( a b)( a b)
a ( a b ) b( a b )
a 2 ab ba b 2
a 2ab b
2
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a ab ba b
2
a 2ab b
2
2
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2
2
5
We know that:
( a b) ( a b)( a b)
2
( a b) 2 a 2 2ab b 2
( a b) 2 1a 2 2ab 1b 2
Observations?
POWERS TOTAL 2 in each TERM
KEY WORDS
We call the expansion
(a b ) n
Binomial
as the original expression has 2 parts.
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1 a 2 2 ab 1 b 2
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1.
2.
3.
4.
5.
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0
(a+b)
1
(a+b)
2
(a+b)
3
(a+b)
4
(a+b)
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( a b)
3
( a b)( a b) 2
( a b)( a 2 2ab b 2 )
a (a 2 2ab b2 ) b(a 2 2ab b 2 )
a 3 2a 2b ab2 a 2b 2ab2 b3
Collecting like terms gives:
( a b)3 a 3 3a 2b 3ab2 b3
Observations?
We again see that there is a clear pattern in the TERMS.
SYMMETRY
POWERS TOTAL 3 in each TERM
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1a 3 3 a 2 b 3 ab 2 1 b 3
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1a 3 3 a 2 b 3 ab 2 1 b 3
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( a b)
4
( a b)( a b)3
( a b)(a 3 3a 2b 3ab 2 b3 )
a (a 3 3a 2b 3ab2 b3 ) b(a 3 3a 2b 3ab2 b3 )
a 4 3a 3b 3a 2b2 ab3 a 3b 3a 2b2 3ab3 b4
Collecting like terms gives:
( a b) 4 a 4 4a 3b 6a 2b2 4ab3 b 4
Observations?
SYMMETRY
POWERS TOTAL 4 in each TERM
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1a 4 4a 3b 6a 2b2 4ab3 1b 4
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1.
2.
3.
4.
5.
0 =1
(a+b)
1
(a+b) =1a+1b
2
2
2
(a+b) =1a +2ab+1b
3
3
2
2
3
(a+b) =1a +3a b+3ab +1b
4
4
3
2 2
3
4
(a+b) =1a +4a b+6a b +4ab +1b
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Powers of
a+b
Pascal’s Triangle
Expression
(a b ) 0
(a b )
1
1
1
(a b ) 2
1
(a b ) 3
(a b ) 4
1
1
1
2
3
4
1
3
6
1
4
1
The numbers of each term in
the expansion match
PASCAL’s TRIANGLE
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PREDICTION?
1, 5, 10, 10, 5, 1
( a b)
5
a 5a b 10a b 10a b 5ab b
5
4
3 2
2 3
4
5
1, 6, 15, 20, 15, 6, 1
( a b) a 6 6a 5b 15a 4b2 20a 3b3 15a 2b 4 6ab5 b6
6
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EXPAND
( x 2)
Power of 0
Power of 1
3
Power of 2
Power of 3
Power of 4
( a b)
3
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1a 3 3 a 2 b 3 ab 2 1 b 3
here a is replaced with x
and b is replaced with 2
( x 2) 1x 3x (2) 3x (2) 1(2)
3
3
2
2
3
( x 2) 1x 3x (2) 3x (4) 1(8)
3
3
( x 2) x 6 x 12 x 8
3
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2
3
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2
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EXPAND
( 3 x 2)
Power of 0
Power of 1
3
Power of 2
Power of 3
Power of 4
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
(3x 2) 3x 3 3x 2 3 3 x 2 2
3
3
2
2
3
(3x 2)3 3x 3 18 x 2 36 x 8
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1. WJEC January 2006 C1 PAST PAPER QUESTION
( a b)3 a 3 3a 2b 3ab2 b3
We start with PASCAL’s
b is replaced with (2)
Here a is replaced with (3x)
The use of brackets will help avoid errors .
(3x 2) (3x ) 3(3 x ) (2) 3(3 x)(2) (2)
3
3
2
2
3
(3x 2) 27 x 3(9 x )(2) 3(3 x )4 8
3
3
2
27 x 54 x 36 x 8
3
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Power of 0
4
(3+2x)
Power of 1
Power of 2
Power of 3
Power of 4
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1 1
1 2 1
1 3 3 1
1 4 6 4 1
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2. WJEC May 2011 C1 PAST PAPER QUESTION
( a b) 4 a 4 4a 3b 6a 2b 2 4ab3 b 4
Here a is replaced with 3
b is replaced with (2x)
The importance of the brackets
must be stressed
(3 2 x ) 4 34 4(3)3 (2 x ) 6(3) 2 (2 x) 2 4(3)(2 x) 3 (2 x) 4
(3 2 x ) 4 81 216 x 216 x 2 96 x 3 16 x 4
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3. WJEC Jan 2010 C1 PAST PAPER QUESTION Jan 2010
We KNOW n=5 and so we can use PASCALS TRIANGLE if we wish.
(a b)5 a5 5a 4b 10a3b2 10a 2b3 5ab4 b5
( a 3x )5 a 5 5a 4 (3x ) 10a 3 (3x ) 2 10a 2 (3 x ) 3 5a (3 x ) 4 (3 x ) 5
TERM in x
TERM in x squared
15a 4 x
COEFFICIENT of the TERM in x
90a 3 x 2
15a 4
COEFFICIENT of the TERM in x squared
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90a 3
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15a 4 x 90a 3 x 2
Solve this equation…..
90a 8 15a
3
90a
a
3
8 15a
3
a
4
3
4
This is a taste of future stories!
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3. WJEC January 2012 C1
Introducing Fractions
(a b)4 a4 4a3b 6a 2b2 4ab3 b4
3
b is replaced with
x
Here a is replaced with x
x
4
2
3
3
3 3
4
3 3
2 3
x 4x 6x 4x
x
x
x
x x
4
2
3
2
3
12
x
6
x
9 4 x 27 81
4
4
2
x x
2
3
x
x
x
x
x
4
x
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4
3
108 81
4
2
x 12 x 54 2 4
x
x
x
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4. WJEC May 2009 C1
4
2
32 16
4
2
x x 8 x 24 2 4
x
x
x
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5. WJEC JANUARY 2009 C1
( a b)5 a 5 5a 4b 10a 3b 2 10a 2b3 5ab 4 b5
Here a is replaced with 1/4
b is replaced with (2x)
BUT!
We are not interested in all terms. We must think carefully, which TERM will involve x cubed?
This is the TERM that will involve x cubed
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5. WJEC JANUARY 2009 C1
( a b)5 a 5 5a 4b 10a 3b 2 10a 2b3 5ab 4 b5
Here a is replaced with 1/4
b is replaced with (2x)
2
BUT!
The TERM containing x cubed is: 10 1 2 x 3
think carefully, which TERM will involve x cubed?
We are not interested in all terms. We must
4
10
3
Which simplifies to: 8 x
16 involve x cubed
This is the TERM that will
10
3
8
x
2
16
5x 3
So the COEFFICIENT of x cubed is
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WJEC May 2012 C1 PAST PAPER QUESTION
First consider the expansion of
( a b) 6
Using PASCAL’S TRIANGLE
a 6a b 15a b 20a b 15a b 6ab b
6
5
Here a is replaced with 1
4 2
3 3
2 4
5
6
b is replaced with (-2x)
(1 2 x ) 1 6( 2 x ) 15( 2 x ) 20( 2 x ) ..
6
2
3
(1 2 x )6 1 12 x 60 x 2 160 x 3 ..
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WJEC May 2007 C1 PAST PAPER QUESTION
(a b)5 a5 5a 4b 10a3b2 10a 2b3 5ab4 b5
2
3
4
1 5
1 1
5
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21
( x ) x 5 x 10 x 10 x 5 x
x
x
x
x
x x
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5
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WJEC January 2013 C1
a6 6a5b 15a 4b2 20a3b3 15a 2b4 6ab5 b6
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WJEC January 2005 C1
(a b)4 a 4 4a3b 6a2b2 4ab3 b4
( a 2 x ) 4 a 4 4a 3 (2 x ) 6a 2 (2 x ) 2 4a (2 x )3 (2 x ) 4
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WJEC January 2007 C1
(a b)4 a4 4a3b 6a 2b2 4ab3 b4
(2 x ) 4 24 4(2)3 x 6(2) 2 x 2 4(2) x 3 x 4
(2 x ) 4 16 32 x 24 x 2 8 x 3 x 4
But we are told this equals… (given in question) So solve the equation
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MAY 2010
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WJEC January 2011 C1
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The Story Continues…..
5 Factorial
5! 5 4 3 2 1
n Factorial
n ! n (n 1) (n 2) ...... 2 1
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PASCAL DEFINED THE NUMBER OF WAYS OF
SELECTING r OBJECTS FROM
n OBJECTS AS:
Sometimes written as
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June 2009 part b
(1 x )n 1 nx
n ( n 1) 2
x
2!
n ( n 1)
2!
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n(n 1) 2 n(n 1)(n 2) 3 n(n 1)(n 2)(n 3) 4
x
x
x ....... x n
2!
3!
4!
Is the TERM in x squared
Is the COEFFICIENT of x squared
n(n 1)
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2!
SOLVE this
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n(n 1)
4 2 2n
2!
Coefficient of x squared is twice the coefficient of x
n2 n
2 4n
1
n 2 n 2n
n 2 3n 0
n( n 3) 0
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So n=3
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JUNE 2007 part b
(1 x )n 1 nx
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n(n 1) 2 n(n 1)(n 2) 3 n(n 1)(n 2)(n 3) 4
x
x
x ....... x n
2!
3!
4!
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JAN 2012 part b
(1 x )n 1 nx
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n(n 1) 2 n(n 1)(n 2) 3 n(n 1)(n 2)(n 3) 4
x
x
x ....... x n
2!
3!
4!
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