AQA Core 4 Algebra
Section 3: The binomial expansion
Notes and Examples
These notes contain subsections on
The general binomial expansion
Harder examples
Finding approximations
Using partial fractions
The general binomial expansion
In Core 2, you came across the binomial expansion which can be used to expand
n
the expression 1 x , where n is a positive integer:
1 x n 1 nx
n(n 1) 2 n(n 1)(n 2) 3
x
x ...
2!
3!
When n is a positive integer, this expansion always has a finite number of terms,
since eventually there will be a factor of 0 in the numerator.
However, this expansion can also be used for any value of n, including negative
numbers and fractions. In cases where n is not a positive integer, there will never be
a zero coefficient, so the expansion will have an infinite number of terms.
This means that the expansion is only valid for cases where values of x r decrease
as r increases, i.e. for -1 < x < 1. Otherwise, the terms of the expansion would get
bigger and bigger and the sum of the terms would increase without limit.
Example 1
Expand (1 x)3 up to the term in x, stating the values of x for which the expansion is valid.
Solution
3 4 2 3 4 5 3
x
x ...
1 2
1 2 3
1 3x 6 x2 10 x3 ...
(1 x)3 1 3x
The expansion is valid for -1 < x < 1.
The expansion still works when the second term in the bracket is a multiple of x. You
just need to replace x in the expansion with the multiple of x, and remember that, for
example, (2x) is not 2x but 8x.
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Also, be careful with the range of x for which the expansion is valid in such cases.
Look carefully at Example 2 below.
Example 2
Expand 1 2x up to the term in x, stating the values of x for which the expansion is valid.
Solution
1 2 x 1 2 x 2
1
1
12
12 32
(2 x)2 2
(2 x)3 ...
1 2
1 2 3
1 x 18 4 x 2 161 8x3 ...
1 12 2 x
1
2
1 x 12 x 2 12 x3 ...
The expansion is valid for
i.e.
-1 < 2x < 1
12 x 12 .
When the second term involves a negative, be very careful with signs.
Example 3
Find the first four terms in the expansion of
1
, stating the values of x for which the
1 x
expansion is valid.
Solution
1
1
1 x
1 x
1 2
1 2 3
x 2
x 3 ...
1 2
1 2 3
2
3
1 x x x ...
1 1 x
The expansion is valid for -1 < x < 1.
Harder examples
You can expand expressions like a b by taking out a factor a n first like this:
n
b
a b a 1 .
a
You can then expand the expression in the bracket using the binomial expansion. Be
very careful it is easy to make mistakes in this kind of work.
n
This method is shown in the next example.
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Example 4
Find the first four terms in the expansion of
expansion is valid.
4 x , stating the values of x for which the
Solution
1
x 2
x 2
4 x 4 x 4 1 2 1
4
4
1
2
3
1 x 12 12 x 12 12 32 x
x 2
1 4 1 2 4 1 2 4 1 2 3 4 ...
2
3
x 1 x 1 x
1 ...
8 8 16 16 64
1
2
1
2
x x2
x3
1
...
8 128 1024
1
x x2
x3
x 2
2 1 2 1
...
4
8 128 1024
x x 2 x3
2
...
4 64 512
The expansion is valid for
i.e.
1 14 x 1
4 x 4
The next example shows a more complicated expansion.
Example 5
Find the first three terms of the expansion of
1 2x
, stating the values of x for which the
1 x
expansion is valid.
Solution
1
1 2x
1
1 2 x 2 1 x 2
1 x
1
1
1
1 2 x 2 1 12 2 x 2 2 (2 x)2 ...
1 2
1 x 18 4 x 2 ...
1 x 12 x 2 ...
1 x
1
2
12 23
( x)2 ...
1 2
3 2
1
1 2 x 8 x ...
1 12 x
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1 2 x 1 x
1
2
1
2
1 x 2 x 2 ...1 12 x 83 x 2 ...
11 12 x 83 x 2 x 1 12 x 12 x 2 1 ...
1 12 x 83 x 2 x 12 x 2 12 x 2 ...
When multiplying out the
brackets, ignore any terms in
powers of x greater than x.
1 x x ...
3
2
3
8
The expansion for 1 2x 2 is valid for
1
i.e.
The expansion for 1 x
12
1 2x 1
12 x 12
1 x 1
is valid for
For both conditions to be true, then x must satisfy the condition 12 x 12 .
You can see further examples using the Flash resource The binomial expansion.
For some practice in finding the range of validity for binomial expansions, try the
binomial puzzle.
Finding approximations
The binomial expansion can be used for finding the approximate value of a function,
by substituting an appropriate value for x and evaluating the first few terms of the
expansion. The more terms are used, the better the approximation.
Example 6
(i) Find the first four terms of the binomial expansion of
(ii) Use your result from (i) to find the value of
(8 x) .
8.1 correct to six decimal places.
Solution
(8 x) 8 x 3 83 1 8x 3 2 1 8x 3
1
(i)
1 x 13 32 x 13 32 53 x
x 3
1
8
8 ...
3 8
2! 8
3!
2
3
x
x
x
1
...
24 576 41472
1
x
x2
x3
x 3
2 1 2 1
...
8
24 576 41472
x x2
x3
...
12 288 20736
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(ii) Let x = 0.1
3
(8 0.1 2
0.1 0.12
0.13
...
12 288 20736
0.1
2.0083333
12
0.1 0.12
3
8.1 2
2.0082986
Using the first three terms
12 288
0.1 0.12
0.13
3
8.1 2
2.0082987
Using the first four terms
12 288 20736
Adding the fourth term does not affect the sixth decimal place
so 3 8.1 2.008299 correct to six decimal places.
3
Using the first two terms
8.1 2
Using partial fractions
The following example shows how partial fractions can be used to simplify the
working if you want to find a binomial expansion.
For example, if you want to find the binomial expansion of
1
up to the
( x 1)(2 x)
term in x, you could write it as x 1 2 x , apply the binomial expansion to each
bracket separately and then multiply out the brackets, discarding any terms of order
greater than x. However, Example 7 shows how using partial fractions can make the
work much easier, as in this case you have to add two algebraic expressions rather
than multiplying.
1
Example 7
1
up to the term in x, stating the range of values of x for which the
( x 1)(2 x)
expansion is valid.
Expand
Solution
1
A
B
( x 1)(2 x) x 1 2 x
Multiplying through by (x 1)(2 x):
Substituting x = 2:
1=B
Substituting x = 1:
1=A
1
1
1
( x 1)(2 x) x 1 2 x
x 1 2 x
1
Both parts must be written in
the form
1 x 21 1 2x
1
1 A(2 x) B( x 1)
1 kx1 before the
expansion can be carried out.
1
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(1)(2)
(1)(2)(3)
( x)2
( x)3 ...
1 2
1 2 3
2
3
1 x x x ...
This expansion is valid for -1 < x < 1.
1 x 1 1 (1)( x)
Expand the two parts
separately
(1)(2) x 2 (1)(2)(3) x 3
2 1 2 3 2 ...
1 2
1 12 x 14 x2 18 x3 ...
This expansion is valid for 1 2x 1 2 x 2
1 2x
1 (1) 2x
1
1
1
1 x 12 1 2x
( x 1)(2 x)
1 x x 2 x3 ... 12 1 12 x 14 x 2 81 x3 ...
1 x x2 x3 12 14 x 81 x 2 161 x3 ...
15 3
12 43 x 87 x 2 16
x ...
The expansion is valid for
values of x for which both parts
are valid, i.e. both -1 < x < 1
and -2 < x < 2 must be true.
The expansion is valid for -1 < x < 1.
Note: another application of partial fractions is in integration. This is covered in the
Differentiation and integration topic.
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