THE BINOMIAL THEOREM
1. Binomial Expansions
A binomial is an expression of the form (a  b)n . Consider the expansion of these binomials for
various values of n.
(a  b)0 
1
1
(a  b)1 
a b
(a  b) 
a  2ab  b
(a  b) 
a  3a b  3ab  b
2
2
3
3
(a  b) 
2
1
2
1
2
3
1
2
1
1
3
3
1
1
4
6
4
1
a  4a b  6a b  4ab  b
Notice the powers of a descend as the powers of b ascend, and the sum of the powers is always n.
The triangle of numbers is known as Pascal’s triangle : the sum of each pair of adjacent numbers
gives the number underneath the pair. The numbers in Pascal’s triangle correspond to the
coefficients in the binomial expansions.
4
4
3
2 2
3
4
Example 1 : Use Pascal’s triangle to expand (2  3x)5 .
We need the next row in the table, and this has coefficients 1, 5, 10, 10, 5, 1.
1
=
=
32
1 32
25
(3 x)1
5
=
=
5 16  3x
240x
24
10
23
(3x)2
=
10  8  9x 2
=
720x 2
10
22
(3x)3
=
10  4  27x3
=
1080x 3
5
21
(3x)4
=
5  2  81x 4
=
810x 4
(3x)5
=
1 243x 5
=
243x5
1
And so (2  3x)5  32  240 x  720 x 2  1080 x3  810 x 4  243x5 .
4
1

Example 2 : Use Pascal’s triangle to expand  5x   .
x

We need the row with coefficients 1, 4, 6, 4, 1.
1
(5 x)4
4
(5 x)3
 1
 
 x
6
(5 x)2
 1
 
 x
4
1
1
4
(5 x)
=
1
x
=
625x 4
=
500x 2
=
4  125x 3  
=
6  25x 2 
1
x2
=
150
=
4  5x  
1
x3
=

=
1
2
 1
 
 x
 1
 
 x
1 625x 4
3
4
1
20 1

And so  5 x    625 x 4  500 x 2  150  2  4 .
x
x
x

1
x4
=
20
x2
1
x4
Example 3 : Find the coefficient of x 3 in the expansion of (7  2 x)5 .
Here we do not need to find the full expansion, just the term in x 3 .
1
5
10
(2 x)3 = 10  49  8x3 = 3920x 3
10
72
5
1
3
The coefficient of x is −3920.
Example 4 : Find the coefficient of x 5 in the expansion of (3x  1)7 .
We need to continue Pascal’s triangle down a few rows.
1
7
21 (3x)5
= 21 243x5 1
12
=
5103x5
The coefficient of x 5 is 5103.
Example 5 : Find the coefficient of x 2 in the expansion of (1  2 x)(3x  5)3 .
We first expand (3x  5)3 …
1
(3x)3
3
(3x)2
3
(3 x)1
1
=
1 27x 3
=
27x 3
(5)1
=
3  9 x 2  5
=
135x 2
(5)2
=
3 3x  25
=
225x
(5)3
=
1 125
=
125
And then multiply by (1  2 x) , only bothering to find those terms involving x 2 .
27x 3
135x 2
225x
125
1
135x
2x
450x
2
The coefficient of x is −585.
Note that if we were far-sighted enough, we would only bother finding the terms in x and x 2 in
the expansion of (3x  5)3 .
2
C2 p72 Ex 5A
2. Formulas for Binomial Coefficients
The numbers in Pascal’s triangle are called binomial coefficients. In most cases, it is not too
much trouble to write the triangle out as far as you need, but for large powers – the value of n in
(a  b)n – this would be time-consuming. So we will now obtain formulas for binomial
coefficients.
But first, some notation. Consider the term involving b r in the expansion of (a  b)n . We use the
n
notation   , alternatively denoted nCr in probability theory, to identify the coefficient, and so
r
n
this particular term is   a n  r b r .
r
The table below shows Pascal’s triangle rewritten with n and r.
r
0
1
2
3
4
0
1
1
1
1
2
1
2
1
n
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
6
1
6
15
20
15
6
 4
5
So for example,    15 ,    1 ,    5 .
 2
0
 4
5
6
1
6
1
The formula for the binomial coefficients is...
n
n!
 
 r  r ! n  r  !
For example,
 7  7!
 
 2  2!5!
7  6  5  4  3  2 1
2  1 5  4  3  2  1
 21

All scientific calculators have an nCr button. On a graphic calculator, to get 7C2 , type 7 OPTN
F6 F3 (to get PROB) F3 (to get nCr) 2 EXE.
C2 p73 Ex 5B
We can write binomial expansions in this notation.
 3
 3
 3
 3
(a  b)3    a3    a 2b    ab 2    b3
0
1
 2
 3
5
 5
 5
 5
 5
 5
(a  b)5    a 5    a 4b    a 3b 2    a 2b3    ab 4    b5
0
1
 2
 3
 4
 5
In general we have the binomial theorem...
n
n
n
 n  n 1  n  n
(a  b) n    a n    a n 1b    a n 2b 2  ...  
 ab    b
0
1
 2
 n  1
 n
Example 1 : Use the binomial theorem to expand (3  2 x)5 .
5
 5
5
 5
 5
 5
(3  2 x)5    35    34 (2 x)    33 (2 x) 2    32 (2 x) 3    3(2 x) 4    (2 x) 5
0
1
 2
 3
 4
 5
2
3
4
5
 1 243  5  162 x  10  108 x  10  72 x  5  48 x  32 x
 243  810 x  1080 x 2  720 x 3  240 x 4  32 x 5
You can get the coefficients from your calculator or straight from Pascal’s triangle if you prefer.
4
1 

Example 2 : Expand  2x  2  .
x 

4
2
3
4
 4
1   4
1   4
1   4

 1   4 1 
4
3
2
 2 x  2    0  (2 x)   1  (2 x)   2    2  (2 x)   2    3  (2 x)   2    4    2 
x   

 x   
 x   
 x    x 
 
4
2
1
 116 x 4  4  8 x  6  2  4  5  1 8
x
x
x
24 8 1
 16 x 4  32 x  2  5  8
x
x
x
Example 3 : Find the term in x 3 in the expansion of (2  3x)8
The required term is
8 5
3
3
  2 (3 x)  56  32  27 x
 3
 48384 x3
Example 4 : Find the term in x 4 in the expansion of (7 x  2)9
The required term is
9
4
5
4
  (7 x) (2)  126  2401x  32
5
 9680832 x3
6
1 

Example 5 : Find the constant term in the expansion of  5x  2  .
x 

1
to the power 2.
x2
You may be able to see that we need to raise 5x to the power 4 and
So the required term is
6
1
4 1 
4
  (5 x)  2   15  625 x  4
x
x 
 2
 9375
2

Example 6 : Find the term in x8 in the expansion of x  x
Here, we need to raise x to the power 4 and

12
.
x to the power 8.
So the required term is
12  4
 x
8
 x
8
 495  x 4  x 4
 495 x8
Also useful is a special case of the binomial theorem...
n
n
n
(1  x) n  1    x    x 2    x3  ...  x n
1
 2
 3
…since all the powers of 1 are all worth 1.



4
Example 7 : Use the binomial theorem to expand 1  x . Hence write 1  3
simplest form.

1 x

4
 4
 1  
1
 4
x  
 2
 
 
2
x
 4
 
 3
   
3
x
 1  4 x  6x  4x x  x2
1  3 
4
 1  4 3  6  3  4  3 3  32
 28  16 3
C2 p75 Ex 5C Q1-6, p76 Ex 5D Q1

x
4

4
in its
3. Approximations Using the Binomial Theorem
If successive terms of a binomial expansion get smaller and smaller, we can ignore negligible
terms and hence make approximations.
Example 1 : Expand (1  2 x)5 up to and including the term in x 3 . Hence find an approximation
for 1.025 .
 5
5
 5
(1  2 x)5  1    (2 x)    (2 x) 2    (2 x)3  ...
1
 2
 3
 1  10 x  40 x 2  80 x3  ...
Substituting x  0.01,
(1.02)5  1  10  0.01  40  0.0001  80  0.000001  ...
 1.10408
This compares favourably with the true value of 1.1040808032
Example 2 : Find the expansion of (2  3x)10 up to and including the term in x 3 . Hence find an
approximation for 1.9710 .
10 
10 
10 
10 
(2  3x)10    210    29 (3 x)    28 (3 x) 2    27 (3 x)3  ...
0
1
2
3
 1024  15360 x  103680 x 2  414720 x3  ...
Substituting x  0.01,
1.9710  1024  15360  0.01  103680  0.0001  414720  0.000001  ...
 880.35328
This agrees with the true value to one decimal place.
Example 3 : If x is so small that terms of x 3 and higher can be ignored, show that
( x  2)(3  2 x)5  486  1377 x  1350 x 2 .
Expanding the binomial,
(3  2 x)5  1 35  5  34 (2 x)  10  33 (2 x) 2
 243  810 x  1080 x 2
We then multiply by ( x  2) , only bothering to find those terms involving x 2 or lower.
x
2
243
810x
1080x 2
243x
486
810x 2
1620x
2160x 2
The expansion is therefore 486  1377 x  1350 x 2 .
The graphs on the next page show that this is a reasonable approximation for values of x close to
zero. The solid line is y  ( x  2)(3  2 x)5 and the dotted line is y  486  1377 x  1350 x 2 .
y
2000
1500
1000
500
– 1
– 0.8
– 0.6
– 0.4
– 0.2
0.2
0.4
0.6
0.8
1
x
C2 p75 Ex 5C Q7-8, p76 Ex 5D Q2,3,5
4. Problems Where the Power is Unknown
We can write...
n
n!
1
 
 0  0!n !
n
n!
n
 
 1  1!(n  1)!
n
n!
n(n  1)

 
2!
 2  2!(n  2)!
n
n!
n(n  1)(n  2)

 
3!
 3  3!( n  3)!
...and so on. This is useful when the power of the binomial is unknown, as in the following
examples.
Example 1 : In the expansion of (1  3x)n , the coefficient of x 2 is 105. Find the value of n.
n
2
2
  (3 x)  105 x
 2
n(n  1)
 9  105
2!
n(n  1)  30
n 2  n  30  0
(n  6)(n  5)  0
n6
We reject the negative root (for now – we will accept it when we get to unit C4!)
Example 2 : In the expansion of (1  px) q , the coefficients of x and x 2 are –28 and 336
respectively. Find the values of p and q.
(1  px) q  1  q( px) 
q (q  1)
( px) 2  ...
2!
Comparing coefficients, we have
pq  28
q (q  1) 2
p  336
2!
Rearranging the first equation…
28
q
784
p2  2
q
p
…and substituting in the second,
q(q  1) 784
 2  336
2!
q
392(q  1)
 336
q
392q  392  336q
56q  392
q7
p  4
Example 3 : In the expansion of (a  x)(1  x) n , the first two terms are
3 16x . Find the values of a and n, and find the coefficients of x 2 and x 3 .
n(n  1) 2 n(n  1)(n  2) 3


(a  x)(1  x) n  (a  x) 1  nx 
x 
x  ... 
2!
3!


an(n  1)  2  n(n  1) an(n  1)(n  2)  3

 a  (1  an) x   n 

x 
 x ...
2! 
3!

 2!

The first two terms are 3 16x , and so
a3
1  an  16
1  3n  16
n5
Substituting into the formulas above for the coefficients of x 2 and x 3 gives us
3 5 4
coefficient of x 2  5 
 35
2
5 4 3 5 4  3
coefficient of x 3 

 40
2
6
C2 p76 Ex 5D Q4, p77 Ex 5E Q1,6
Topic Review : Binomial Theorem