the coefficients in the binomial expansion of (1 + x ) 5

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.

1 5 10 10 5 1Want to know some BRAIN BLOWING MATHS?

The number of ways of winning the jackpot on the National Lottery is 13 983 816

13 983 816 seconds is 161 days – from 11th April until 19th September.

13 983 816 two pence pieces laid end to end would stretch 220 miles –

from London to Paris.

496 13,983,816C

……………… …………………….

22/04/2023 Mrs Richards 3

The Binomial ExpansionFor Positive Integers

INTRODUCTION

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Lets start at the very beginning!

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STARTER1. What is the value of two to the power of four ?

0( ) 1a b 6. What is any number to the power of zero ?

2. What is the cube of 2x ?

42 16

2 2(5 ) 25y y

3 3(2 ) 8x x

4. Simplify 3 23 (2 ) (5 )x y

NOT32x

3. What is the square of 5y ?

3 2 3 23 8 25 600x y x y

0 1a 7. What is ( a + b ) to the power of zero ?

9. Expand and simplify 2( )a b

8. Expand and simplify 1( )a b a b That’s all we can do

5. What is the cube of2x

3

3

2 8x x


2( ) ( )( )a b a b a b

( ) ( )a a b b a b ( )( )a b a b ONE WAY….. ANOTHER WAY…..

2 2a ab ba b 2 2a ab ba b

2 22a ab b 2 22a ab b


Observations?

POWERS TOTAL 2 in each TERM

TERM COEFFICIENTKEY WORDS

We call the expansion

Binomial as the original expression has 2 parts.

nba )(

2( ) ( )( )a b a b a b We know that:

2 2 2( ) 2a b a ab b 2 2 2( ) 1 2 1a b a ab b


22 baba 1 2 1


Expand these Binomials:

1. (a+b)0

2. (a+b)1

3. (a+b)2

4. (a+b)3

5. (a+b)4

Mrs Richards 9

3)( ba

3223 babbaa 331 1

323 2 22 22 aa b aba bb ba

2( )( )a b a b

2 2( )( 2 )a b a ab b

2 2 2 2( 2 ) ( 2 )a a ab b b a ab b

Collecting like terms gives: 2 323 3( ) 3 3a aba bb a b Observations?


SYMMETRY

22/04/2023

We again see that there is a clear pattern in the TERMS.


3223 babbaa 331 1

22/04/2023 Mrs Richards

4)( ba

3 3 44 32 232 233 33 bab aba b aa b a b ba

3( )( )a b a b

3 2 2 3( )( 3 3 )a b a a b ab b

3 2 2 3 3 2 2 3( 3 3 ) ( 3 3 )a a a b ab b b a a b ab b

Collecting like terms gives:

23 42 34 4 6 4) 4( aa b a bb a b ab Observations?


SYMMETRY

11

4 3 2 2 3 41 4 6 4 1a a b a b ab b


Spot anything?

1. (a+b)0

2. (a+b)1

3. (a+b)2

4. (a+b)3

5. (a+b)4

=1

=1a+1b

=1a2+2ab+1b2

=1a3+3a2b+3ab2+1b3

=1a4+4a3b+6a2b2+4ab3+1b4


Powers of a + bPascal’s Triangle

3)( ba

2)( ba

Expression

1 2 1

1 3 3 1

1)( ba 1 1

0)( ba

4)( ba 1 4 6 4 1

1

The numbers of each term inthe expansion match

PASCAL’s TRIANGLE


5)( ba

PREDICTION?1, 5, 10, 10, 5, 1

5 4 3 2 2 3 4 55 10 10 5a a b a b a b ab b

6 5 4 2 3 3 2 4 5 66 15 20 15 6a a b a b a b a b ab b 6( )a b

1, 6, 15, 20, 15, 6, 1


How is this useful?EXPAND 1

1 1 1 2 1 1 3 3 1 1 4 6 4 1

Power of 0

Power of 1

Power of 2

Power of 3

Power of 4

3223 babbaa 331 1 3)( bahere a is replaced with x and b is replaced with 2

3)2( x

3 3 2 2 3( 2) (2) (21 3 3 ) )1(2x x x x

3 3 2( 2) 6 12 8x x x x

3 3 2( 2) (2)1 3 3 1(4) (8)x x x x


Slightly harder?EXPAND 1

1 1 1 2 1 1 3 3 1 1 4 6 4 1

Power of 0

Power of 1

Power of 2

Power of 3

Power of 4

3)23( x

3 3 2 2 3(3 2) 3 3 3 2 3 3 2 2x x x x

3 3 2(3 2) 3 18 36 8x x x x RUBBISH!


1. WJEC January 2006 C1 PAST PAPER QUESTION

3 3 2 2 3(3 2) (3 ) 3(3 ) (2) 3(3 )(2) (2)x x x x

3 3 2

3 2

(3 2) 27 3(9 )(2) 3(3 )4 8

27 54 36 8

x x x x

x x x

3 3 2 2 3( ) 3 3a b a a b ab b We start with PASCAL’s

Here a is replaced with (3x) b is replaced with (2)

The use of brackets will help avoid errors .


Try for yourself

(3+2x)4 1

1 1 1 2 1 1 3 3 1 1 4 6 4 1

Power of 0

Power of 1

Power of 2

Power of 3

Power of 4


2. WJEC May 2011 C1 PAST PAPER QUESTION

4 4 3 2 2 3 4( ) 4 6 4a b a a b a b ab b

Here a is replaced with 3 b is replaced with (2x)

The importance of the brackets must be stressed

4 4 3 2 2 3 4(3 2 ) 3 4(3) (2 ) 6(3) (2 ) 4(3)(2 ) (2 )x x x x x

4 2 3 4(3 2 ) 81 216 216 96 16x x x x x


3. WJEC Jan 2010 C1 PAST PAPER QUESTION Jan 2010

We KNOW n=5 and so we can use PASCALS TRIANGLE if we wish.

5 5 4 3 2 2 3 4 5( ) 5 10 10 5a b a a b a b a b ab b

5 5 4 3 2 2 3 4 5( 3 ) 5 (3 ) 10 (3 ) 10 (3 ) 5 (3 ) (3 )a x a a x a x a x a x x

TERM in x TERM in x squared

COEFFICIENT of the TERM in x

COEFFICIENT of the TERM in x squared

4 3 215 90a x a x

415a390a


COEFFICIENTS

TERMS 4 3 215 90a x a x

48 15a390a

This is a taste of future stories!

Solve this equation…..3

3

908 15

a aa

34

a


3. WJEC January 2012 C1

4 4 3 2 2 3 4( ) 4 6 4a b a a b a b ab b

Introducing Fractions

Here a is replaced with x b is replaced with

3x

4 2 3 44 3 23 3 3 3 34 6 4x x x x x

x x x x x

4 3 24

2 3 4

3 12 6 9 4 27 81x x xx xx x x x x

44 2

2 4

3 108 8112 54x x xx x x

2

2


44 2

2 4

2 32 168 24x x xx x x

4. WJEC May 2009 C1


5. WJEC JANUARY 2009 C1

5 5 4 3 2 2 3 4 5( ) 5 10 10 5a b a a b a b a b ab b

BUT! We are not interested in all terms. We must think carefully, which TERM will involve x cubed?

Here a is replaced with 1/4 b is replaced with (2x)

This is the TERM that will involve x cubed


5. WJEC JANUARY 2009 C1

5 5 4 3 2 2 3 4 5( ) 5 10 10 5a b a a b a b a b ab b

BUT! We are not interested in all terms. We must think carefully, which TERM will involve x cubed?

Here a is replaced with 1/4 b is replaced with (2x)

This is the TERM that will involve x cubed

The TERM containing x cubed is: 2

3110 24

x

Which simplifies to: 310 816

x

310 816

x

35x

2

So the COEFFICIENT of x cubed is 5


WJEC May 2012 C1 PAST PAPER QUESTION

First consider the expansion of 6( )a b Using PASCAL’S TRIANGLE

6 5 4 2 3 3 2 4 5 66 15 20 15 6a a b a b a b a b ab b

Here a is replaced with 1 b is replaced with (-2x)

6 2 3(1 2 ) 1 6( 2 ) 15( 2 ) 20( 2 ) ..x x x x

6 2 3(1 2 ) 1 12 60 160 ..x x x x


WJEC May 2007 C1 PAST PAPER QUESTION

5 5 4 3 2 2 3 4 5( ) 5 10 10 5a b a a b a b a b ab b

2 3 4 55 5 4 3 21 1 1 1 1 1( ) 5 10 10 5x x x x x x

x x x x x x


WJEC January 2013 C1

6 5 4 2 3 3 2 4 5 66 15 20 15 6a a b a b a b a b ab b



23 42 34 4 6 4) 4( aa b a bb a b ab

3 42 24 34 4 (24 (2 ) 6( (2 )2 ) (2) )aa x xaa x xxa



4 4 3 2 2 3 4( ) 4 6 4a b a a b a b ab b

4 4 3 2 2 3 4(2 ) 2 4(2) 6(2) 4(2)x x x x x

4 2 3 4(2 ) 16 32 24 8x x x x x

But we are told this equals… (given in question) So solve the equation


MAY 2010


5 Factorial5! 5 4 3 2 1

n Factorial! ( 1) ( 2) ...... 2 1n n n n

The Story Continues…..


PASCAL DEFINED THE NUMBER OF WAYS OF SELECTING r OBJECTS FROM n OBJECTS AS:

Sometimes written as


2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)(1 ) 1 .......2! 3! 4!

n nn n n n n n n n nx nx x x x x

June 2009 part b

2( 1)2!

n n x Is the TERM in x squared

( 1)2!

n n Is the COEFFICIENT of x squared

( 1) 552!

n n SOLVE this


( 1) 4 2 22!

n n n

Coefficient of x squared is twice the coefficient of x2

2

2

2 41

2

3 0( 3) 0

n n n

n n n

n nn n

So n=3


JUNE 2007 part b

2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)(1 ) 1 .......2! 3! 4!



2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)(1 ) 1 .......2! 3! 4!


JAN 2012 part b

the coefficients in the binomial expansion of (1 + x ) 5

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