the following slides show one of the 55 presentations that cover the a2 mathematics core modules c3...

The following slides show one of the 55 presentations that cover the A2 Mathematics core modules C3

and C4.

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““Teach A Level Maths”Teach A Level Maths”Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules

44: More 44: More Binomial Binomial ExpansionsExpansions

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

More Binomial Expansions

Module C4

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In AS we developed a formula to expand expressions of the formnx)1( as a finite series of terms provided that n was a positive integer.The formula is

nnnnn xxCxCCx ...)1( 2210

!!

!

128

20

!!

!

)( rnr

nC r

n

wher

e

820Ce.g

.


We are now going to extend the work to expansions of

nx)1(

where n is negative and/or a fraction.

We will also find out how to expand expressions of the form

nbxa )(

e.g. ,1)1( x ,21

)4( x 3)2( x


Let’s look again at where n is a positive integer

nx)1(

822

81

80

88 ...)1( xxCxCCx

We’ll take n = 8 as an example.

!!

!

)( rnr

nC r

n

Using

we get

828

08

8...

62

8

71

8

80

8)1( xxxx

!!

!

!!

!

!!

!

!!

!

Instead of using a calculator to work out these coefficients, we can simplify them.


12345123

12345678

828

08

8...

62

8

71

8

80

8)1( xxxx

!!

!

!!

!

!!

!

!!

!

e.g.!!

!

80

81

1

!!

!

71

8

12345671

12345678

!!

!

62

812345612

12345678

!!

!

53

8

( since is defined as 1 )

!0

1

8

12

78

123

678


123

678

!5!3

!83

8

C

123

)2)(1(

)!(!3

!3

nnn

rn

nCn

( Notice that there are the same number of factors in the numerator and denominator. )

If we replace 8 by n


We can replace each coefficient of the binomial expansion in a similar way.

So,nnnnn xxCxCCx ...)1( 2

210

becomes

32

123

)2)(1(

12

)1(1)1( x

nnnx

nnxnx n

nx...It can be shown that this expansion is true even when n is not a positive integer BUT with 2 properties• The series becomes infinite

11 x( We say it only converges if )

• It is only valid for 11 x

More Binomial ExpansionsSUMMAR

Y...

123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

• The series is infinite

• The series only converges to if

11 x

nx)1(

( also written as )1xN.B. The notation cannot be

used. rnC

• n can be a positive or negative fraction or a negative integer.( If n is a positive integer the series is

finite. )


xx )1(1)1( 1

We aren’t going to prove the Binomial expansion but I can illustrate it in 2 ways.

1nConsider

...123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

1x

1)1( 1 x

I simplify each term by deciding the sign first, then the constants ( by cancelling ), and the x’s last.

2

12

)2)(1(x ...

123

)3)(2)(1( 3

x


xx )1(1)1( 1


x

1)1( 1 x


1nConsider

...123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

1x

2

12

)2)(1(x ...

123

)3)(2)(1( 3

x


2

12

)2)(1(x

xx )1(1)1( 1


x1)1( 1 x 2x


1nConsider

...123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

1x

...123

)3)(2)(1( 3

x


...123

)3)(2)(1( 3

x


x1)1( 1 x 2x 3x


1nConsider

...123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

1x

xx )1(1)1( 1

2

12

)2)(1(x



xx )1(1)1( 1

x1)1( 1 x 2x 3x ...


1nConsider

...123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

1x

2

12

)2)(1(x ...

123

)3)(2)(1( 3

x


...1)1( 321 xxxx

Using the binomial theorem, we have

Now the l.h.s. equals , sox1

1

...11

1 32

xxxx

The r.h.s. is an infinite geometric progression with a = 1 and r = 1, so,

)(1

1

xS

r

aS

1

x

1

1... shl


...11

1 32

xxxx

We can also draw the graphs of the l.h.s. and the r.h.s. of our expression.

Drawing both graphs, we’ll start with

and, as a first approximation, the first 2 terms

of the

xy

1

1

xy 1polynomial on the r.h.s., .


xy

1

1


Remember that the expansion is only valid for

1xbut even so it is not good.

xy 1

xy

1

1

More Binomial ExpansionsWe will now include one more term of the

polynomial at a time and draw the new graphs.

As we include more terms in the polynomial

the graph should get closer to the graph of x

y

1

1

So we will have x

y

1

1

21 xxy

and xy 1

321 xxxy etc.


21 xxy

xy

1

1


321 xxxy

xy

1

1


4321 xxxxy

xy

1

1


54321 xxxxxy

xy

1

1


Also, the graphs show very clearly that the expansion is not valid outside these values.

115432 ...1 xxxxxxy

The fit is now very good between x = 1 and x = 1

xy

1

1

Although we haven’t proved the binomial expansion, we have seen that it works in this example.


In AS Maths we saw how to expand binomial series when n was a positive integer.

e.g. If n = 4,

Our new version of the formula works for positive integers also, as we eventually reach a term that is zero. All the terms after this will also be zero, so the series is finite ( as before ).

We will now apply the formula to expressions where n is not a positive integer.

012345

))(3)(2)(1( 5

xnnnn 4n

More Binomial Expansionse.g.2 Find the 1st 4 terms in ascending

powers of x of the following. For each one, give the values of x for which the expansion is valid.(i) (ii

)2)1( x 2

1

)1( x (iii) 21

)21( x

Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by)1( x2


Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by ( 2 ))1( x2

(i) (ii)

2)1( x 21

)1( x (iii) 21

)21( x

e.g.2 Find the 1st 4 terms in ascending powers of x of the following. For each one, give the values of x for which the expansion is valid.


Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by

replace x by

( 2 ))1( x2

(i) (ii)

2)1( x 21

)1( x (iii) 21

)21( x



Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by

replace x by( x )( 2 ))1( x

2

(i) (ii)

2)1( x 21

)1( x (iii) 21

)21( x



Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by ( 2 ))1( x2

(i) (ii)

2)1( x 21

)1( x (iii) 21

)21( x


replace x by( x )

More Binomial ExpansionsSolution: (i)

1)1( 2x

The brackets are essential !

...12312

1)1(32

nx )1( nnnx x )2)(1( nnn x

For , replace n by


2


...12312

1)1(32

nx

Solution: (i)

1)1( 2x


)1( nn x )2)(1( nnn x

For , replace n by


nx

)2(

2


...12312

1)1(32

nx

Solution: (i)

1)1( 2x )2(


)1( nnnx x )2)(1( nnn x

For , replace n by


2

)( x


...12312

1)1(32

nx

Solution: (i)

1)1( 2x )2(


I go straight to (3) here . . .

)1( nnnx x )2)(1( nnn x

For , replace n by


2

)( x )2( )3(

( since (21) spreads the expression out too much ),BUT it is the only simplification I make at this stage.


...12312

1)1(32

nx

Solution: (i)

1)1( 2x )2(


)1( nnnx x )2)(1( nnn x

For , replace n by


2

)( x )2(12)3( )( x 2


...12312

1)1(32

nx

Solution: (i)

1)1( 2x )2(


)1( nnnx x )2)(1( nnn x

For , replace n by

( x )( 2 ))1( x

2

)( x )2(12)3( )( x 2

)2( )3(

)4(

replace x by


...12312

1)1(32

nx

Solution: (i)

1)1( 2x )2(


)1( nnnx x )2)(1( nnn x

For , replace n by


2

)( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(


As before, we simplify in the order:

signs, numbers, letters

1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21 Tip: Don’t use a calculator to simplify the

terms. It’s slow, fiddly and prone to error.

( You will have to practise doing it without and be systematic ! )


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

There are 4 minus signs here NOT 3

signsnumber

sletters


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

3


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

2x3


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

2x How many minus signs here?

ANS: 6

3


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

2x 43


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

2x 4 3x3


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21

signsnumber

sletters

2x 4 3x3

Valid for 1xN.B. It’s convenient to write the denominators

as factorials. If you do this, take care with the cancelling. I’ll do the next example this way.


(ii)

21

)1( x

...12312

1)1(32


To avoid piled up fractions, we can put the factorial under the

2x

Replace n by2

1

1)1( 21

x

2

1x

2

1

2

1

!2

2x


(ii)

21

)1( x

...12312

1)1(32


1)1( 21

x

2

1x

2

1

2

1

!2

2x

2

1

2

1

2

3

!3

3x

1 8

2x 16

23x

2

x

Replace n by2

1

Valid for 1x


16821)1(

3221 xxx

x

(iii) 21

)21( xAs we’ve just worked out the expansion for

21

)1( xwe can obtain the new expansion by replacing x by ( 2x ) in that answer.

1)21( 21

x2

)2( x8

)2( 2x

16

)2( 3x

1 x 8

4 2x

2

16

8 3x

2

221

32 xxx Valid for 12 x 2

1 x

More Binomial ExpansionsExercis

esWrite down the 1st 4 terms, in the expansions in ascending powers of x, for the following and give the values of x for which the expansions are valid.

1. 3)1( x 41

)1( x 31

)31(

x2. 3.


...12312

1)1(32


Replace n by ( 3 ) and x by ( x )

1)1( 3x

!2

))(4)(3( 2x

!3

))(5)(4)(3( 3x

x31 26x 310x

2

2

2

3)1( xSolutions: 1.

))(3( x

Valid for 1x


...12312

1)1(32


1)1( 41

x!24

3

4

1 2x

!34

7

4

3

4

1 3x

41

x

32

3 2x

128

7 3x

41

)1( xSolutions: 2.

x

4

1

Replace n by )( 41

2

Valid for 1x


...12312

1)1(32


1)31( 31

x!2

)3(

3

4

3

1 2x

!3

)3(

3

7

3

4

3

1 3x

x1 22x3

14 3x

)3(3

1x

Replace n by and x by ( 3x )

)( 31

3

Solutions: 3. 31

)31(

x

2

2

31 x

Valid for 13 x


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


In AS we developed a formula to expand expressions of the formnx)1( as a finite series of terms provided that n was a positive integer.

The method can be extended to expressions of the form

nx)1( where n is negative and/or a fraction.

nbxa )( and

More Binomial ExpansionsSUMMAR

Y...

123

)2)(1(

12

)1(1)1( 32

xnnn

xnn

xnx n

• The series is infinite

• The series only converges to if

11 x

nx)1(

( also written as )1xN.B. The notation cannot be

used. rnC

• n can be a positive or negative fraction or a negative integer.( If n is a positive integer the series is

finite. )

More Binomial Expansionse.g.2 Find the 1st 4 terms in ascending

powers of x of the following. For each one, give the values of x is the expansion valid.

Solution: (i)

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n

For , replace n by)1( x2

(i) (ii)

2)1( x 21

)1( x (iii) 21

)21( x

replace x by( x )( 2 )

1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(


1)1( 2x )2( )( x )2(12)3( )( x 2

)2(123

)3( )( x 3

)4(

x21 2x 4 3x3

Valid for 1xN.B. It’s convenient to write the denominators

as factorials. If you do this, take care with the cancelling.

signsnumber

sletters

Simplify in the order:


(ii)

21

)1( x

1)1( 21

x

2

1x

2

1

2

1

!2

2x

2

1

2

1

2

3

!3

3x

1 8

2x 16

23x

2

x

Replace n by2

1

Valid for 1x

...123

)2)(1(

12

)1(1)1(

32

xnnnxnn

nxx n


16821)1(

2221 xxx

x

(iii) 21

)21( xAs we’ve just worked out the expansion for

21

)1( xwe can obtain the new expansion by replacing x by ( 2x ) in that answer.

1)21( 21

x2

)2( x8

)2( 2x

16

)2( 3x

1 x 8

4 2x

2

16

8 3x

2

221

32 xxx Valid for 12 x 2

1 x

Full version available from:- Chartwell-Yorke Ltd.

114 High Street, Belmont Village,

Bolton, Lancashire,

BL7 8AL England, tel (+44) (0)1204 811001, fax

(+44) (0)1204 811008

[email protected]

mailto:[email protected]

http://www.chartwellyorke.com/

the following slides show one of the 55 presentations that cover the a2 mathematics core modules c3...

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