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Binomial Series Summary1. Expansion of [ ]nxf )(1+

This is given by

( )[ ]

( )[ ]

( )[ ] ............)(

!4

)3)(2)(1()(

!3

)2)(1()(

!2

)1()(1

432+

+

+

++ xf

nnnnxf

nnnxf

nnxnf

Example: ( ) .............)3(!3

)4)(3)(2()3(

!2

)3)(2()2)(3(131 32

2+

+

++=

xxxx

.............1082761 32 ++++= xxx

Note:

(i) For ,n ,+n the structure intended for expansion must strictly adhere to the form

[ ]nxf )(1+ . For instance, ( )31

28 x+ must be reorganized as3

1

3

1

418

+

xat the very beginning

before proceeding with the actual binomial series expansion.

(ii) The expansion of [ ]nxf )(1+ is valid for .1|)(|

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Example: if we are aware that ,8

1

2

11

1

1 241

xxx

xy +

+

=

Then xxxdx

d

dx

dy

4

1

2

1

8

1

2

11 2 +=

+

Hence, at the point ,2

1,

3

14

gradient of the tangent to the curve

4

1

1

1

+

=

x

xy is approximately

equal to12

5

3

1

4

1

2

1=

+ , and the equation of this tangent is simply given by

,3

1

12

5

2

14

= xy ie

++=

4 2

1

36

5

12

5xy (shown)

3. Finding the formula of the general term in the expansion

Example: In the expansion of

22

22

41

16

1)4(

+=+

xx , coefficient of nx2

( ) ( ) nnn

n

n

n

n

+

=

= 4

1

!

)1)........(4)(3(21

16

1

4

1

!

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

16

1

=( ) ( ) ( )

24

)1(1

)16(4

)1(1

4

1

!

)!1(1

16

1+

+=

+=

+

n

n

n

nnnnn

n

n(shown)

4. Ascending series expansion

Example:2

1

24

11

+

xmay resemble the required format [ ]nxf )(1+ , but if an ascending series is to

be achieved through expansion, then24

1)(

xxf = will not be appropriate as the main variable

resides purely in the denominator. A transformation has to be made:

( ) 21

22

1

2

2

1

2 1441

411

+

=

+ xxx

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( )

+

+

+

+= .......)4(

!3

2

5

2

3

2

1

)4(!2

2

3

2

1

)4(2

112 32222 xxxx