HSC MATHEMATICS: MATHEMATICS EXTENSION 1
3 UNIT MATHEMATICS
TOPIC 17
BINOMIAL THEOREM
n – factorial n!
For positive integers n the product of the first n natural numbers is called
n – factorial and is written
n!  (1)(2)(3)
(n  1)(n)
0!  1 1!  1 2!  (1)(2)  2 7!  (1)(2)(3)(4)(5)(6)(7)  5040
A binomial is a polynomial of two terms such as 2 x  5 y or 2  (a  b)3 .
The Binomial Theorem is a quick way of expanding a binomial expression that
has been raised to some power n , for example,  5 x  6  .
12
ONLINE: 3 UNIT MATHS
1
Consider the polynomial ( x  y)n where n is an integer n  1, 2, 3,
This
polynomial can be expanded using the Binomial Theorem in terms of the
n
binomial coefficients nCk   
n  1,2, 3,
0k n
k 
n
 k  is read as ‘n over k’
 
n
n
n!
Ck    
 k  k ! n  k !
n
4
4
n
C0     1
0
 4
C0     1
0
0
4
0k n
 0
C0     1
 0
n
n
n
 n 
n!
Ck     n Cn  k  


k 
 n  k  k ! n  k !
 n
Cn     1
 n
 4  (4)(3)(2)(1)
C1    
 4  4C3
1  (1)(3)(2)(1)
 4  (4)(3)(2)(1)
C3    
 4  4C1
 3  (1)(2)(3)(1)
4
4
 4  (4)(3)(2)(1)
C2    
6
 2  (1)(2)(2)(1)
 4
C4     1
 4
ONLINE: 3 UNIT MATHS
2
Pascal’s triangle
An easy way to calculate the binomial coefficients is to use Pascal’s Triangle
where the binomial coefficients are arranged in a triangle. Each interior number
is the sum of the nearest two numbers in the previous row.
ONLINE: 3 UNIT MATHS
3
6
 6  (6)(5)(4)(3)(2)(1)
 6  (6)(5)(4)(3)(2)(1)

1


6
 15
6
5
 4 
 
  (5)(4)(3)(2)(1)(1)
  (4)(3)(2)(1)(1)(2)
 6  (6)(5)(4)(3)( 2)(1)
 6  (6)(5)(4)(3)(2)(1)


20
 15
3
 2 
  (3)(2)(1)(1)(2)(3)
  (2)(1)(1)(2)(3)(4)
 6  (6)(5)(4)(3)(2)(1)
6


6
1 
0 1
  (1)(1)(2)(3)( 4)(5)
 
The construction of a Pascal Triangle is based upon the relationship between
binomial coefficients
n
Ck  n1 Ck 1  n1 Ck
1  k  n 1
This is known as Pascal’s Triangle Identity.
Example
LHS  6C2  15 RHS 5 C1 5 C2  5  10  15  LHS  RHS
ONLINE: 3 UNIT MATHS
4
n
Proof
n
Ck 
k 1
Ck  n1 Ck 1  n1 Ck
n!
k !  n  k !
n
C1 
1  k  n 1
1 k  n
n!
n
1!  n  1!
n 1
C0  n1 C1  1 
 n  1!  1  n  1  n
1!  n  2 !
QED
k2
n 1
 n  1!   n  1!
 k  1!  n  k !  k !  n  k !
 n  1!  1  1    n  1!  k  n  k 

 k  1!  n  k !  n  k k   k  1!  n  k !  (k )(n  k ) 
Ck 1  n1 Ck 

n!
 n Ck
k !  n  k !
QED
The coefficients of the variables x and y in the expansion of  x  y  are called the binomial
n
coefficients. The (k+1)th binomial coefficient of order n ( n a positive integer) is
n
n
n
n!
Ck    
 k  k !  n  k !
(k+1)th binomial coefficient
n
Ck    gives the number of combinations of n things k at a time.
k 
ONLINE: 3 UNIT MATHS
5
Binomial Theorem
n
1 
( x  y)  x    x
n
n
n 1
n
y    x n2 y 2 
2
 n  n1

x y  yn

 n  1
 n  nk k k  n n
( x  y)     x
y   Ck x n  k y k
k 0  k 
k 0
k n
n
Example Expand the binomial expression (a  b)6
 Construct a Pascal Triangle to find the binomial coefficients
(a  b)6  a6  6a5 b  15a 4 b2  20a3 b3  15a 2 b4  6ab5  b6

The sum of the binomial coefficients is equal to 2n and is obtained by setting
x  y 1
n6
(1  1)  2 
6
6
k n
n
k 0
 
 k 
ONLINE: 3 UNIT MATHS
 1  6  15  20  15  6  1  64
6
The binomial theorem is proved by induction. That is, it is shown to hold for n = 1, and further
shown that if it holds for any given value of n then it also holds for the next higher value of n.
 n  nk k
( x  y)     x
y
k
k 0  
n
Suppose ( x  y )     x nk y k is true. Now consider
k 0  k 
k n
k n
n
 x  y
n 1
n
  x  y  x  y 
n
n
n
  x  y     x nk y k
k 0  k 
n
n
n
n
    x n1k y k     x n j y j 1
k 0  k 
j 0  j 
n
n 1 n
n
n
n
 
   x n1     x n1k y k    y n1     x n j y j 1
k 1  k 
j 0  j 
0
n
Let j  k  1 k  j  1
 x  y
n 1
x
n 1
y
n 1
 n  n1k k n  n  n1k k
   x
y  
y
x
k 1  k 
k 1  k  1 
n
 n   n   n1k k
     
y
 x
k 1  k 
 k  1 
 n  1  n   n 
Using Pascal's Identity 
   k    k  1
k

   

n
n 1 n  1
 n  1 n1k k

 n1k k
n 1
x
y

y
 x  y   x n1  y n1   



x
k
k
k 1 
k 0 


x
n 1
y
n 1
n
ONLINE: 3 UNIT MATHS
QED
7
Example (a) Use the binomial theorem to expand  2  3 x  .
5
(b) What is the 4th term in the expansion for increasing powers of x?
(c) Find the largest coefficient in the expansion.
(d) Expand
2  3 x
5
(e) Use the binomial theorem to differentiate the function y   2  3 x 
5
(f) Use the binomial theorem to integrate y   2  3 x  from x = 0 to x = 1.
5
 (a)
Always write down the binomial theorem to start answering the question
( x  y) 
 5  nk k

 x y
k 0 k 
2  3 x
5 
5 k
k
     2   3 x k
k 0 k 
k 5
5
5
k 5
5 
Use Pascal’s triangle to give the binomial coefficients    1 5 10 10 5 1
k 
2  3 x
5
 (1)(25 )(1)  (5)(24 )(3) x  (10)(23 )(32 ) x 2  (10)(22 )(33 ) x3  (5)(2)(34 ) x 4  (1)(35 ) x5
 32  240 x  720 x 2  1080 x3  810 x 4  243x5
ONLINE: 3 UNIT MATHS
8
(b) increasing powers of x:
4th term
x0
x1
x2
x3
x4
5
5 3
3
t4     2   3 x3  1080 x 3
3
x5
4th term  k  3
in agreement with part (a)
(c) Let the coefficients be represented by ak . Since x and y are positive, we can check whether
the coefficients are increasing or decreasing by considering the ratio ak 1 / ak
If ak 1 / ak  1
ak  2 / ak 1  1
ak  ak 1  ak 2
0k n
 ak 1 is the largest coefficient
 5  5k 1 k 1
 3   5  k   3 
 k  1  2


ak 1 / ak 

   1  k  2.6 k  2 k  1  3
k  1  2 
 5  5 k k

 k   2  3 
 
Therefore a3  1080 is the largest coefficient in agreement with part (a).
 5  nk k
( x  y)     x
y
k 0 k 
k 5
5
(d)
2  3 x
5
2  3 x
5
5 
5 k
k
     2  (1) k  3 x k
k 0 k 
k 5
 (1)(25 )(1)  (5)(24 )(3) x  (10)(23 )(32 ) x 2  (10)(22 )(33 ) x3  (5)(2)(34 ) x 4  (1)(35 ) x5
 32  240 x  720 x 2  1080 x3  810 x 4  243x5
ONLINE: 3 UNIT MATHS
9
(e)
y  2  3 x
5
dy / dx  (5)(3)(2  3 x) 4
 4  nk k

 x y
k 0 k 
k 4
( x  y) 
4
Pascal's triangle
n  4 1 4 6 4 1
dy / dx  (15)  24  (4)(2)3 (3 x)  (6)(2) 2 (3 x) 2  (4)(2)(3 x)3  (3 x) 4 
dy / dx  240  1440 x  3240 x 2  3240 x3  1215x 4
y   2  3 x   32  240 x  720 x 2  1080 x3  810 x 4  243x5
5
dy / dx  240  1440 x  3240 x 2  3240 x3  1215 x 4
QED
(f)
y  2  3 x
5
I    2  3 x  dx
1
5
0
6 1
 1 
1
I     2  3x       56  26   864.5
 0  18 
 18  
y   2  3 x   32  240 x  720 x 2  1080 x 3  810 x 4  243x5
5
I    32  240 x  720 x 2  1080 x3  810 x 4  243x5  dx
1
0
1
I  32 x  120 x 2  240 x3  270 x 4  162 x5  40.5 x 6 
0
I  32  120  240  270  162  40.5  864.5 QED

ONLINE: 3 UNIT MATHS
10
1

Find the coefficients of x and x in the expansion of  2x  
x

4
Example
6
3
Check your answer by expanding the expression using the binomial theorem.

 6  6 k k

 x y
k 0 k 
k 6
( x  y) 
6
Pascal's triangle
n  6  1 6 15 20 15 6 1
k 6
6
1  k  6  6  6 k 6 k
k k
k
6 k

2
x


2
x

1
x

 
 1  2   x62 k 



    


x  k 0  k 

k 0 k 
6
For the term in x 4 we require 6  2k  4  k  1, therefore, the coefficient ak 1 is
6 5
ak 1     2   1  (6)(32)  192
1 
For the term in x 3 we require 6  2k  3  k  1.5 , k is not an integer, there is no term in x 3
6
1

6
4
2
2
4
6
 2 x    64 x  192 x  240 x  160  60 / x  12 / x  1 / x
x


ONLINE: 3 UNIT MATHS
11
Example Find the term in the expansion of  3  5 x  with the greatest coefficient.
20

Always write down the binomial theorem to start answering the question
 20  nk k

k  x y
k 0 

k  20
( x  y)
20

3  5 x  
20
 
  k   3  5 
k  20
k 0
20

20  k

k
xk
In the expansion of  x  y  where x and y are both positive, then successive coefficients in the
n
expansion get larger and then smaller. Therefore, the ratio R of the (k+1)th coefficient to the kth
coefficient will exceed one or be equal to one until the largest term is reached.
If ak 1 / ak  1
ak  2 / ak 1  1
ak  ak 1  ak 2
 ak 1 is the largest coefficient
20!
320k 1 5k 1
(k  1)!  20  k  1!
 5  1 
R
  
  20  k   1
20!
3
k

1
20 k
k



3
5
k !  20  k !
100  5 k  3k  3 8k  97 k  12.125  k  12 k  1  13
 20 
The largest term is term is   37 513 x13
13 

ONLINE: 3 UNIT MATHS
12
Example
(a) Expand  x  1
n
(b) Expand  x  1
substitute x  1
n
substitute x  1 hence show that
n
2n1  nC1  n C3  n C5 
(c) Show that
n1
(d) Prove
C1  n C3  n C5 
Ck n Ck  n Ck 1
 2n  n  n 
(e) Show that      
 n  k 0  k 
 nC0  n C2  n C4 
 nC0  n C2  n C4 
which gives the Pascal’s triangle identity
2

Always write down the binomial theorem to start answering the question
k n
( x  y) 
n
n
 k  x
k 0
nk
yk
 
 n  nk

k  x
k 0  
k n
(a)
(b)
( x  1) 
n
( x  1) 
n
k n
n
k 0
 
  k   1
k n
x 1

2 
n
k 0
k
n
 k 
 
n
C0  n C1  n C2 
 n Cn
x nk
ONLINE: 3 UNIT MATHS
13
n
k
n
0      1 n C0 n C1  n C2  nC3   1 nCn
k 0  k 
n
C1  n C3  n C5   nC0  n C2  n C4 
k n
x 1
(c)
Add the results from parts (a) and (b)
2 2
n

n
 x  1
 x  1

C0  n C2  n C4 
2n1  n C0  n C2  n C4 
(d)

n 1
n 1

 nC1  n C3  n C5 
n 1 n  1

 n1k
 
x
k
k 0 

 n  1
coefficient for x n1k is 

 k 
n
n
n
n
n
n
n
 ( x  1)  x  1  ( x  1)   x nk     x n1k     x nk
k 0  k 
k 0  k 
k 0  k 
n  n 
coefficient for x n1k is    

 k   k  1
Therefore,
n1
Ck n Ck  n Ck 1
ONLINE: 3 UNIT MATHS
n n
 k   Ck
 
14
 2n  2 n  k

 x
k 0  k 
k  2n
(e)
( x  1)

2n
 2n 
We are interested in the term   x n when k  n .
n 
k n
n
Also, ( x  1) n     x nk
k 0  k 
( x  1) ( x  1) 
n
n
 n  n k

k  x
k 0  
k n
k n
n
k 0
 
 k  x
nk
The term in x n from the expansion expressed as a product is
n
 n 
k 0
 
  k  x
nk
n
 k 
n  k  x 



x n is formed by taking the product of terms in x nk from the first expression and x k from
the second expression in the product.
n  n 
k   n  k 
  

 n  n  k

  x
k  0  k 
n
n
 k  n  n  2  n
 n  k  x     k   x


 k  0   
Hence, equating the coefficients of the terms for x n
 2n  n  n 
 n    k 
  k 0  
2

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