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The following slides show one of the 55 presentations that cover the A2 Mathematics core modules C3
and C4.
Demo DiscDemo Disc
““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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More Binomial Expansions
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
.
More Binomial Expansions
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
More Binomial Expansions
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.
More Binomial Expansions
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
More Binomial Expansions
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
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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. )
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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
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xx )1(1)1( 1
I simplify each term by deciding the sign first, then the constants ( by cancelling ), and the x’s last.
x
1)1( 1 x
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
2
12
)2)(1(x ...
123
)3)(2)(1( 3
x
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2
12
)2)(1(x
xx )1(1)1( 1
I simplify each term by deciding the sign first, then the constants ( by cancelling ), and the x’s last.
x1)1( 1 x 2x
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
...123
)3)(2)(1( 3
x
More Binomial Expansions
...123
)3)(2)(1( 3
x
I simplify each term by deciding the sign first, then the constants ( by cancelling ), and the x’s last.
x1)1( 1 x 2x 3x
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
xx )1(1)1( 1
2
12
)2)(1(x
More Binomial Expansions
I simplify each term by deciding the sign first, then the constants ( by cancelling ), and the x’s last.
xx )1(1)1( 1
x1)1( 1 x 2x 3x ...
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
2
12
)2)(1(x ...
123
)3)(2)(1( 3
x
More Binomial Expansions
...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
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...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., .
More Binomial Expansions
xy
1
1
More Binomial Expansions
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.
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21 xxy
xy
1
1
More Binomial Expansions
321 xxxy
xy
1
1
More Binomial Expansions
4321 xxxxy
xy
1
1
More Binomial Expansions
54321 xxxxxy
xy
1
1
More Binomial Expansions
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.
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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
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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.
More Binomial Expansions
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
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.
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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
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.
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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.
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
replace x by( x )( 2 ))1( x
2
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x
The brackets are essential !
)1( nn x )2)(1( nnn x
For , replace n by
replace x by( x )( 2 ))1( x
nx
)2(
2
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x )2(
The brackets are essential !
)1( nnnx x )2)(1( nnn x
For , replace n by
replace x by( x )( 2 ))1( x
2
)( x
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x )2(
The brackets are essential !
I go straight to (3) here . . .
)1( nnnx x )2)(1( nnn x
For , replace n by
replace x by( x )( 2 ))1( x
2
)( x )2( )3(
( since (21) spreads the expression out too much ),BUT it is the only simplification I make at this stage.
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x )2(
The brackets are essential !
)1( nnnx x )2)(1( nnn x
For , replace n by
replace x by( x )( 2 ))1( x
2
)( x )2(12)3( )( x 2
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x )2(
The brackets are essential !
)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
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...12312
1)1(32
nx
Solution: (i)
1)1( 2x )2(
The brackets are essential !
)1( nnnx x )2)(1( nnn x
For , replace n by
replace x by( x )( 2 ))1( x
2
)( x )2(12)3( )( x 2
)2(123
)3( )( x 3
)4(
More Binomial Expansions
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 ! )
More Binomial Expansions
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
More Binomial Expansions
1)1( 2x )2( )( x )2(12)3( )( x 2
)2(123
)3( )( x 3
)4(
x21
signsnumber
sletters
3
More Binomial Expansions
1)1( 2x )2( )( x )2(12)3( )( x 2
)2(123
)3( )( x 3
)4(
x21
signsnumber
sletters
2x3
More Binomial Expansions
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
More Binomial Expansions
1)1( 2x )2( )( x )2(12)3( )( x 2
)2(123
)3( )( x 3
)4(
x21
signsnumber
sletters
2x 43
More Binomial Expansions
1)1( 2x )2( )( x )2(12)3( )( x 2
)2(123
)3( )( x 3
)4(
x21
signsnumber
sletters
2x 4 3x3
More Binomial Expansions
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.
More Binomial Expansions
(ii)
21
)1( x
...12312
1)1(32
nx )1( nnnx x )2)(1( nnn x
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
More Binomial Expansions
(ii)
21
)1( x
...12312
1)1(32
nx )1( nnnx x )2)(1( nnn 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
More Binomial Expansions
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.
More Binomial Expansions
...12312
1)1(32
nx )1( nnnx x )2)(1( nnn x
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
More Binomial Expansions
...12312
1)1(32
nx )1( nnnx x )2)(1( nnn x
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
More Binomial Expansions
...12312
1)1(32
nx )1( nnnx x )2)(1( nnn x
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
More Binomial Expansions
More Binomial Expansions
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.
More Binomial Expansions
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(
More Binomial Expansions
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:
More Binomial Expansions
(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
More Binomial Expansions
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
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