Binomial Expansion
Pg 1
x 2  5x
A
B
C
. Express f(x) in the form
+
+
,
2
1 x
1 x
(1  x) (1  x)
(1  x) 2
where A, B and C are constants. The expansion of f (x), in ascending powers of x, is
r
2
3
C0 + C1x + C2 x + C3 x + ……..+ Cr x +…….
Let f (x) =
Find C0 , C1 , C2 and show that C3 = 11. Express Cr in terms of r.
2
2
x + 5x ≡ A (1  x) + B (1 + x)(1  x) + C (1 + x)
2
Put x = −1 ,
(−1) + 5 (−1) = A [ 1 − (−1) ]
2
 A =
2
Comparing coefficient of x : A  B = 1
Comparing constant
: A+B+C = 0
∴ f (x) = 
 (1 + x)
 2 (1  x)
3 (1  x)
1
1
2
=
2
3
r r
2
3
x
 [ 1  x + x  x + ……... + (1) x + ……….]
=
r
2 [ 1 + x + x + x + ……... +
+ ……….]
(2) (3)
(2) (3) (4) 3
2
(x ) 
x + ……...
2!
3!
r
(2) (3)(4).........(r  1)
+
(x) + ……….]
r!
=
3 [ 1 + (2)(x) +
=
3 [ 1 + 2x + 3x + 4x + ……... + (r + 1) x + ……….]
2
r
=
=
=
=
 1  B = 1
 1  2 + C = 0
1
2
3
−
+
1 x
1 x
(1  x) 2
r +1
∴ Coefficient of x in f (x) = (1)
Hence C0
C1
C2
C3
(1) 2  5(1)
[1  (1)]2
1
(1)
2
(1)
3
(1)
4
(1)




2
2
2
2
+
+
+
+
3 (1)
3 (1 + 1)
3 (2 + 1)
3 (3 + 1)
r
3
 2 + 3 (r + 1)
=
=
=
=
0
5
6
11
(Shown)
= 1
 B = 2
 C = 3
Binomial Expansion
Pg 2
2
(a) Find the term independent of x in the expansion of (x +
2 6
).
x
2
1
(b) Find the coefficient of x6 in the expansion in ascending powers of x of (4 + x ) 2 .
2 6
) ,
x
2
(a) General term of (x +
6
Tr +1 =
6r
C r (x2 )
2 r
x
( )
6
=
6
∴ The term independent of x =
1
(b) (4 +
12  2r
r r
·2 x
12  3r = 0
For independent term,
2
x )2
C rx
6
=
r
12  3r
C r· 2 ·x
 r= 4
4
C2 · 2
= 240
1
2
= 2 (1 + 14 x ) 2
2
= 2 [ 1 + 12 ( 14 x ) +
6
∴ Coefficient of x
Express ( 4 + x
1
)2
=
where
1
2
( 12 )( 12  1) 1 2 2
( 12 )( 12  1)( 12  2) 1 2 3
(4x ) +
( 4 x ) + ……]
2!
3!
( 12 )( 12  1)( 12  2) 1 3
(4 )
3!
1
[x(
=
1
x2[
=
x 2 (1 +
[ x (1 +
4 2
) ]
x
( 12 )( 12 ) 4 2
( 12 )( 12 )( 32 ) 4 3
( ) +
( ) + …..…]
x
x
2!
3!
2
 22 + 43 + ……..)
x
x
x
n
Note that if x is large, we have to expand (a + x)
(or descending powers of x).
1
up to the first four terms.
x
1
=
4
1 + 12 ( ) +
x
1
1
512
x > 4 as a series of ascending powers of
4
+ 1) 2 ]
x
(4+x)2 =
=
1
=
x2 +
2
1
x2

2
3
x2
+
as a series of ascending powers of
4
5
x2
1
x
+ ……..
Binomial Expansion
Pg 3
2
1
1
+

and that x is small.
2x  1
x 1
(x  1) 2
Find the series expansion of f(x) in ascending powers of x as far as the terms in x3. Write down the
range of values of x for which the expansion is valid.
Given that f (x) =
2
1
+
2x  1
(x  1) 2
= 2 (1  2x)
1
1
x 1

+ (1 + x)
2
2
3
2
3
=
1
2 (2x  1)
2
1
 (x + 1)
+ (x + 1)
1
 (1 + x)
2
3
2
3
2
3
2
3
= 2 [ 1 + 2x + (2x) + (2x) + ……] + [ 1  2x + 3x  4x + ……]  [ 1  x + x  x + …….]
(2  4x  8x  16x + …..)
=
2
+
(1  2x + 3x  4x + …….) 
(1  x + x  x + …….)
3
=  2  5x  6x  19x  ……..
Valid for 2x < 1 &

x <1
x < 12 . i.e.  12 < x < 12 .
1
1
 2
.
x2
x 1
Find the series expansion in descending powers of x of f (x) up to and including the 3rd non-zero
term. Write down the restriction that must be imposed on x for the expansion to be valid.
Given that f (x) =
1
1
 2
x2
x 1
1
1 1
=
x (1  2x )
=
x
=
x
=
x
1
1
1
1
(x  2)
=
2

2
1
2
2 1
1 2
2
+ x
+ 4x
+ 4x
3
3
= [ x (1 − 2x−1 ) ]−1 − [ x2 (1 + x−2 ) ]−1
x (1 + x )
[ 1 + (2x ) + (2x ) + ……]
+ 2x
1
 (x + 1)
+ …….  [ x
2
2
2

x [ 1  (x
4
 x
6
2 2
) + (x )  …….]
+ x
 ……..]

x >2 &
x >1

x > 2.
i.e. x <  2 or x > 2
+ ……..
Valid for 2x 1 < 1 &
x 2 < 1
Binomial Expansion
Pg 4
r
Find the coefficient of x in the expansion, in ascending powders of x , of
1
(1  x) 3
= (1  x)
3
r
(3)(4)
(3)(4)(5)........(r)(r  1)(r  2)
2
(x) + ……. +
(x) + ……..
2!
r!
= 1 + (3)(x) +
Coeff. of x
r
1
.
(1  x) 3
=
r
(3)(4)(5)........(r)(r  1)(r  2)
(1)
r!
=
r
(1) r  3  4  5........(r)(r  1)(r  2)
(1)
r!
=
r! (r  1)(r  2)
(1)
1 2  r!
=
2r
1 2  3  4  5........(r  2)
(1)
1  2  r!
=
(r  1)(r  2)
2
3
3
Find the expansion in ascending powers of x, up to and including the term in x , of (2  x) .
r
Hence, find the coefficient of x .
(2  x)
3
=
3
3
2 (1  12 x)
=
1
(3)(4)(5) 1 3
(3)(4) 1 2
[ 1 + (3)( 12 x) +
( 2 x) +
( 2 x)
8
2!
3!
=
1
5 3
3
3 2
[1 + 2 x + 2 x +
x
8
4
r
Coefficient of x
+ ……]
1
3
3 2
5 3
+
x +
x +
x
8
16
16
32
=
=
r
1 (3)(4)(5)........(r)(r  1)(r  2)
( 12 )
8
r!
=
r
r
1 (1) r  3  4  5........(r)(r  1)(r  2)
(1) ( 12 )
3
r!
2
=
=
1
2
r 3
1
2
r 3
2r
3  4  5........r (r  1) (r  2)
(1)
r!
1  2  3  4  5........r (r  1) (r  2)
1  2  r!
+ ……]
=
(r  1)(r  2)
2 r4
+ ……
Binomial Expansion
Pg 5
Express the following as a series of ascending powers of x, up to and including the term in x3.
(a) (4 +
1
x) 2
3
(b) (x  2)
1
1
(a) (4 + x) 2
2 [ 1 + ( 12 )( x ) +
4
=
2 (1 +
=

=
1
[ 1 + (3)(
8
1
3
 [1 + 2 x +
8

1
1 x)
2
9  3x
=
5 3
3 2
x +
x
2
4


1
3
2!
1
Express (1 + 4x)
(3)(4)(5) 1 3
( 2 x)
3!
+
=

+ ……]
1
3
3 2
5 3

x 
x 
x
8 16
16
32
+ ……]
1
=

1
x) ] 2
3
[ 9 (1 +
( 12 )( 23 ) 1 2
( x)
3
2!
1
1 2
5
3
1
(1  x +
x 
x + ……)
3
6
24
432
+

<1
1
2,
( 12 )( 12 )( 32 ) x 3
( ) + ……]
3!
4
+
x <2
=
( 12 )( x)

4
+ ……]
1
[1 +
3
1x
3
2
3
3
(2) (1  12 x)
=
Valid for
(x)
(3)(4) 1 2
( 2 x)
2!
+
(9 + 3x) 2
=
( 12 )( 12 )
x < 4
3
Valid for  12 x < 1
(c)
1
x
x2
x3
x
x2
x3

+
+ ……..) = 2 +

+
+ ……..
8
128
1024
4
64
512
[ 2 (1  12 x) ]
=
9  3x
4
=
x
3
1
4 2 (1 + x ) 2
4
Valid for 4 < 1
(b) (x  2)
1
[ 4 (1 + x ) ] 2 =
=
(c)
=
+
( 12 )( 23 )( 25 ) 1 3
( x)
3
3!
+ …… ]
1
1 2
5
3
1

x +
x 
x + ……
3
18
72
1296
x <3
where x > 14 as a series of descending powers of x, up to and including the
third non-zero term.

1
1
(1 + 4x) 2 =
[ 4x (1 +
1 2
)] ,
4x
1
( 12 )( 23 ) 1 2
( 12 )( 23 )( 52 ) 1 3
( ) +
( ) + ……]
4x
4x
2!
3!
=
1 x  2 [ 1 + ( 1 )( 1 ) +
2
2 4x
=
1 x2  1 x2 + 3 x2
2
16
256
1
3
5
Binomial Expansion
Pg 6
1
1  px
in ascending powers of x are identical up to and
1  qx
2
including the term in x , find the values of p and q.
Given that the expansions of (1 + x) 5 and
By using this result and putting x = 
1
, find the approximate 5th root of 30 in the form of an
16
improper function.
(1 +
1
x) 5
1  px
1  qx
1 ( 4 )
5
5 x2
=
1
1 + x +
5
=
(1 + px)(1 + qx)
2!
1
+ ……..
=
1 +
1
2 2
x 
x + ……..
5
25
2 2
(1 + px)(1  qx + q x + ……..)
=
2
2
1 + (p  q) x + (q  pq) x + ……
1
1
 p = q + -------------------(1)
Comparing coeff. of x :
pq =
5
5
2
2
2
Comparing coeff. of x : q  pq = 
------------------------------------------------(2)
25
2
1
2
1
2
2
Subst. (1) into (2) :
q  (q + ) q = 
  q = 
 q=
5
25
5
25
5
3
1
1 5x
1
2
3
∴ (1) :
p= +
=
Hence (1 + x) 5 
.
5
5
5
1 2x
=
5
Subst. x = 
 (
1
,
16
(1 
1
16
1
)5
30 15
77
) 
32
78
1

1

5
3 ( 1 )
5
16
2 ( 1 )
5
16
30
2

77
78
 (
15 15
77
) 
16
78

5
30 
1
77
39
2
Give the binomial expansion, for small x, of (1 + x) 4 up to and including the term in x , and simplify
the coefficients. By putting x = 1 in your expression, show that
16
(1 +
1
x) 4
=
Let x = 1 ,
16
 (
( 14 )( 43 ) 2
1
1 + 4x +
x + ………
2!
=
4 17

3
8317
.
4096
1 + 14 x +
x + ………
32
2
1
3 1 2
∴ (1 + 1 ) 4  1 + 14 ( 1 ) +
( ) + ………
32 16
16
16
17 14
8317
) 
16
8192
1
8317
8192
8317
 2(
)
8192
 12 (17) 4 

4 17

4 17

8317
4096
(Shown)
Binomial Expansion
Expand (1 + x)
1
4
Pg 7
2
in ascending powers of x up to and including the term in x .
Use your expansion with x =
(1 + x)
1
4
( 14 )( 45 ) 2
1 + ( 14 ) x +
x + ………
2!
=
1
4
=
2
(1 +
1  14
)
80
=
(
1
81  4
)
80
=
(
1
80 4
)
81
4 80
=
=
4 81
4 (2 4 )(5)
=
3
2 (45)
3
1
into the binomial series,
80
1
5 1 2
= 1  14 ( ) +
( ) + ………
80
32 80
Subst. x =
2 (45)
3
45
5
1  14 x +
x + ………
32
=
1
into LHS :
80
Subst. x =
(1 + x)
1
to find an approximation value for 4 5 correct to 5 decimal places.
80
=
1
1
3
+
)( 2 )
320
40760
(1 
=
1.49535
(to 5 d.p.)
1
2 
2x ) 2
6
Expand (1 
in ascending powers of x as far as the term in x , and state the range of values
of x for which the expansion is valid.
Hence evaluate

1
2
(1  2x ) 2
1
0.98
to 7 decimal places.
1 + ( 12 )(2x ) +
=
1 + x + 2 x + 52 x + ……..
3 4
2
Valid for 2x 2 < 1
Compare (1 
1
2 2
2x )
=
2
∴
0.98
6
x 2 < 12

Let 1  2x = 0.98
1
( 12 )( 23 )
( 12 )( 23 )( 52 )
2 2
2 3
(2x ) +
(2x ) + ……
2!
3!
2
=
1
with
1  2x 2
[12
x
1
<
2
i.e. 
1
2
< x <
1
0.98
 x = 0.1.
1
=

2 
(0.1) ] 2
3
=
1 + (0.1) + 2 (0.1) + 52 (0.1) + ……..
=
1.0101525
2
4
( to 7 d.p.)
6
1
2
Binomial Expansion
Pg 8
1  x 13
) where
1 x
x < 1, simplifying the coefficients. By putting x to a suitable value, use your series to find an
2
Find the expansion in ascending powers of x, up to and including the term in x , of (
approximate value for (
(
1  x 13
) =
1 x
15 13
) , giving your answer in the form of a fraction.
17
1
(1  x) 3 (1 + x)
1
3
1
2
( )( 3 )
1
1
2
[ 1 + (x) + 3
(x ) + …….] [ 1 + ( ) x +
3
3
2!
1
1 2
1
2 2
x + …….]
[ 1  x  x + …….] [1  x +
3
9
3
9
2 2
1  23 x +
x + ……..
9
=
=
=
1 

1 
Subst. x = 1 ,
16
1
1  3
16
1 
16 
2
 1  23 ( 1 ) +
(1 )
9 16
16
2
( 13 )( 53 ) 2
x + ……..]
2!
 (
15 13
1105
) 
17
1152
2
Find the expansion in a series of ascending powers of x, up to and including the term in x , of
1 x
.
1 x
By putting x to a suitable value in your expansion, find 11 correct to 3 decimal places.
1 x
1 x
1
x) 2 (1

 x)
1
2
=
(1 +
=
[ 1 + 12 x +
=
[ 1 + 12 x +
x + …….] [ 1 + 12 x +
x + …….]
8
8
=
1
Let x =
,
10
( 12 )( 12 ) 2
( 12 )( 23 )
2
x + …….] [ 1 + ( 12 )(x) +
(x) + …….]
2!
2!
1
2
2
1 + x + 12 x + ……..
1
1
1
10
1
10
=
1
1 2
11
 1 +
+ 12 ( )
10
10
9
Note : if let
3
2
1 x
= 11
1 x
1 +
1
1 2
+ 12 ( ) + ……..
10
10

11
 1.105
3
 x =
5
6
gives

11 = 3.315
11 = 2.18.
(inaccurate)