Associate Degree 2024 – 2025 First Semester

CCMA4002 Linear Algebra

Chapter 2
Binomial Expansions

2.1 Expansion of (x + y)n when n is a positive integer

A. The Notations n! and Cnr

The product of the first n positive integers is called n factorial and is denoted by n!
n!  n  (n 1)  (n  2)   2 1
where n is a positive integer.

For example, 5!  5  4  3  2 1  120

8!  8  7  6  5  4  3 2 1  40320

In particular, 0! is defined as 1, i.e., 0! = 1.

If we choose r objects from n different objects and the order of the objects being chosen
is unimportant, then we call the situation a combination of the objects.
The number of ways, i.e., the number of combinations, of r objects taken from n different
objects is denoted by the notation C nr

n! n(n  1)(n  2) (n  r  1)
Cnr  
r!(n  r)! r!

where n and r are non-negative integers and r  n .

6! 6! 6  5  4  3  2 1 65
For example, C62      15
2!(6  2)! 2!4! (2 1)  (4  3  2 1) 2 1

8! 8! 8  7  6  5  4  3  2 1 8 7  6
C83      56
3!(8  3)! 3!5! (3  2 1)  (5  4  3  2 1) 3  2 1

In particular, C 0n  C nn  1

2-1
B. Binomial Theorem

n
(x  y) n   C nr x n  r y r
r 0

 C x n  C1n x n 1 y  C2n x n  2 y 2   C rn x n  r y r   C nn 1xy n 1  C nn y n

n
0

n(n  1) n  2 2 n(n  1)(n  2) (n  r  1) n  r r

 x n  nx n 1 y  x y   x y   nxy n 1  y n
2! r!

This is called the Binomial Theorem for a positive integral index.

In the expansion (x + y)n ,

(1) There are n + 1 terms, beginning with xn and ending with yn.

(2) The powers of x are in descending order while those of y are in ascending order.
The sum of the powers of x and y in each term is equal to n.

(3) The (r + 1)th term, Cnr x n r yr , is called the general term of the expansion of (x  y)n in the
descending powers of x.

(4) The coefficients Cnr , r = 0, 1, 2,…, n are called binomial coefficients.

All the binomial coefficients are integers.

2-2
Example 1

Expand (x + 2)6 in descending powers of x.

Solution:

(x  2)6  C60 x 60 (2)0  C16 x 6 1 (2)1  C62 x 6 2 (2) 2  C36 x 6 3 (2)3  C64 x 6 4 (2) 4  C65 x 6 5 (2)5  C66 x 6 6 (2)6
 (1)(x 6 )(1)  (6)(x 5 )(2)  (15)(x 4 )(4)  (20)(x 3 )(8)  (15)(x 2 )(16)  (6)(x)(32)  (1)(x 0 )(64)
 x 6  12x 5  60x 4  160x 3  240x 2  192x  64

Example 2

4
Expand  2x  1  in descending powers of x.
 x

Solution:

4 2 3 4
 1  1  1  1  1
 2x   = (2x)4 + C14 (2x)3    + C42 (2x)2    + C34 (2x)    +   
 x  x  x  x  x

 1  1   1  1
= 16x4 + (4)(8x3)    + (6)(4x2)  2  + (4)(2x)   3  + 4
 x x   x  x
8 1
= 16x4 – 32x2 + 24 – 2
+ 4
x x

Example 3

Expand (1 + x + x2 )4 in ascending powers of x.

Solution:

(1 + x + x2 )4 = [1 + (x + x2)]4
= 14 + C14 13 (x + x2) + C42 12 (x + x2)2 + C34 (1)(x + x2)3 + (x + x2)4
= 1 + (4)(x + x2) + (6)(x2 + 2x3 + x4) + (4)(x3 + 3x4 + 3x5 + x6)
+ (x4 + 4x5 + 6x6 + 4x7 + x8)
= 1 + (4x + 4x2) + (6x2 + 12x3 + 6x4) + (4x3 + 12x4 + 12x5 + 4x6)
+ (x4 + 4x5 + 6x6 + 4x7 + x8)
= 1 + 4x + 10x2 + 16x3 + 19x4 + 16x5 + 10x6 + 4x7 + x8

2-3
C. The Pascal’s Triangle

The numerical coefficients in the binomial expansion can be arranged in the form of a triangle as below:

n Coefficient
0 1 1
1 11 1 1
2 1 C12 1 1
2 1
3 1 C13 C32 1 1 3 3 1
4 1 C14 C42 C34 1 1 4 6 4 1
5 1 C15 C52 C53 C54 1 1 5 10 10 5 1
6 1 C16 C 62 C36 C 64 C56 1 1 6 15 20 15 6 1

The Pascal’s Triangle is symmetric.

In the triangle, the numbers in each row at both ends are 1 and the others are the sums of the two
adjacent numbers in the row above.

Theorem In the Pascal triangle, the binomial coefficient satisfies

(a) Cnr = Cnnr for r = 0, 1, 2,…, n [Symmetric Pascal triangle]
(b) Cnr1 + Cnr = Cnr 1 for r = 1, 2,…, n [Pascal rule]

Proof :
n! n! n!
(a) Cnr = = = = Cnnr
(n  r )!r! r!(n  r )! [n  (n  r )]!(n  r )!

n! n!
(b) Cnr1 + Cnr = +
[n  (r  1)]!(r  1)! (n  r )!r!

n! n!
= +
(n  r  1)!(r  1)! (n  r )!r!

r n!  (n  r  1) n!
= ( ∵ (n – r + 1)! = (n – r + 1)(n – r)!
(n  r  1)!r!
and r! = r (r – 1)! )
n! (r  n  r  1)
=
(n  r  1)!r!

n! (n  1)
=
(n  r  1)!r!

(n  1)!
=
[( n  1)  r )]! r!

= Cnr 1

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D. Finding Specific Terms

Particular terms of an expansion can be obtained by substituting suitable values of r into the general term
instead of finding the whole expansion.

Example 4
9
 3 
In the expansion of  2x  2  , find
 x 

(a) the coefficient of x3 and

(b) the constant term.

Solution:
9
 3 
The general term in the expansion of  2x  2  is
 x 

r
9 r  3  9 9 r 9 r 2 r 9 9r r 9 3r
C (2x)   2   Cr 2 x (3) (x )  Cr 2 (3) x
9 r
r
 x 

(a) Set 9 – 3r = 3, we have r = 2

98
Hence the coefficient of x3 is C92 29-2 (–3)2 =  27  9 = 41472
2

(b) Set 9 – 3r = 0, we have r = 3

98 7
Hence the constant term is C93 29-3 (–3)3 =  26  (–27) = –145152
3  2 1

2-5
Example 5

Let n be a positive integer. In the expansion of (x3 + 3)n in descending powers of x,

the coefficient of the third terms is 189. Find the value of n and the coefficient of x9.

Solution:

The third term in the expansion of (x3 + 3)n is Cn2 (x3)n-2(3)2, i.e., 9 Cn2 x3n-6

9C2  189
n
Therefore,
n(n  1)
9  189
2
9n(n  1)  378
n(n  1)  42
n 2  n  42  0
(n  7)(n  6)  0
n = 7 or n = – 6 (rejected as n is a positive integer)

The general term of the expansion of (x3 + 3)7 is C7r (x3 )7r (3)r  C7r 3r x 213r

Set 21 – 3r = 9, we have r = 4

7  6 5 4
The coefficient of x9 is C7434 =  81 = 2835
4  3  2 1

2-6
2.2 Expansion of (1+x)n when n is NOT a positive integer and x < 1

n (n  1) 2 n (n  1)(n  2) 3 n (n  1)(n  2)  (n  r  1) r
(1 + x)n = 1 + nx + x + x + … + x +…
2! 3! r!

If x < 1, the above formula is valid for any real values of n.

The result is known as the binomial theorem for a fractional or negative index.
n (n  1) 2
Remark : 1. For fractional or negative n, the series 1 + nx + x + … does not terminate,
2!
it is an infinite series.
2. The infinite series is convergent and has (1 + x)n as the limit of its sum
only when x < 1, i.e., –1 < x < 1.
Therefore the expansion is valid only if x is in the range –1 < x < 1 and
this range must always be stated.

3. The expansion is NOT valid for (a + x) n.

n
To expand (a + x)n it must first be written in the form an 1   and
x
 a
x
this expansion is valid only for –1 < a < 1, i.e., –a < x < a.

Example 6
Expand the following expansions in ascending powers of x up to and including the term in x3.
State the range of values of x for which the expansion is valid.
(a) (1  x)1 (b) (1  x)1 (c) (1  2x)1

Solution:
(1)(1  1) 2 (1)(1  1)(1  2) 3
(a) (1  x) 1  1  (1)x  x  x 
2! 3!
 1  x  x 2  x3  for x  1 i.e., 1  x  1

(1)(1  1) (1)(1  1)(1  2)

(b) (1  x) 1  1  (1)( x)  ( x) 2  ( x)3 
2! 3!
 1 x  x  x 
2 3
for x  1 i.e., 1  x  1

OR By replace x by – x in (a)

(1  x)1  1  (x)  (x)2  (x)3   1  x  x 2  x3  for x  1 i.e., 1  x  1

(1)(1  1) (1)(1  1)(1  2)

(c) (1  2x)1  1  (1)(2 x)  (2x) 2  (2x)3 
2! 3!
1 1
 1  2x  4x 2  8x3  for 2x  1 i.e.,   x 
2 2
OR By replace x by 2x in (a)
1 1
(1  2x)1  1  (2x)  (2x)2  (2x)3   1  2x  4x 2  8x 3  for 2x  1 i.e.,   x 
2 2

2-7
Example 7

Expand the following expansions in ascending powers of x up to and including the term in x3.
State the range of values of x for which the expansion is valid.
1 1 1
 
(a) (1  x) 2 (b) (1  x) 2
(c) (4  x) 2

Solution:

 1  1   1  1  1 
   1    1  2 
1
1
(a) (1  x)  1    x   
2 2  2  2  2  2  x3 
2
x 
 
2 2! 3!
1 1 1
 1  x  x 2  x3  for x  1 i.e., 1  x  1
2 8 16

 1  1   1  1  1 
     1      1   2 
 1
1
 1    x  
 2  2  2  2  2  2  x3 
(b) (1  x) 2
x 
 2  2! 3!
1 3 5
 1  x  x 2  x3  for x  1 i.e., 1  x  1
2 8 16

1

 x
1 1
  2
(c) (4  x) 2
4 2
1  
 4
  1  1   1  1  1  
      1 2      1   2  3 
1  1  x   2  2   x   2  2  2  x
 1             
4   2  4  2! 4 3! 4 
 
1 1 3 2 5 3 
 1  x  x  x  
2 8 128 1024 
1 1 3 2 5 x
  x x  x3  for  1 i.e., 4  x  4
2 16 256 2048 4

2-8
2.3 Approximations

Consider (1  x)20 and its binomial expansion,

(20)(19) 2 (20)(19)(18) 3 (20)(19)(18)(17) 4
(1  x)20  1  20x  x  x  x   x 20
2! 3! 4!
This is valid for all values of x.

So if, for example, x = 0.01 we have

(20)(19) (20)(19)(18) (20)(19)(18)(17)
(1  0.01)20  1  20(0.01)  (0.01)2  (0.01)3  (0.01)4   (0.01)20
2! 3! 4!
i.e., (1.01)20  1  0.2  0.019  0.00114  0.00004845  1040

In fact, taking only the first four terms give (1.01)20  1.22014.
This approximation is correct to three decimal places as the fifth and succeeding terms do not add
anything to the first four decimal places.
Therefore, we have (1.01)20 = 1.220 correct to 3 decimal places

In general, if x is small so that successive powers of x quickly become negligible in value,

then the sum of the first few terms in the expansion of (1 + x)n gives an approximate value for (1 + x)n.

Remark : When finding an approximation, the binomial expansion of (1 + x)n and NOT (a + x)n
should be used, e.g. to find the approximate value of (3.006)5 we use 35 (1 + 0.002)5.

2-9
Example 8

By substituting 0.002 for x in the expansion of (1  x)5 , find the value of (3.006)5 correct to
three decimal places.

Solution:

(3.006)5  (3  0.006)5
 35 (1  0.002)5
 35 [15  C15 (0.002)  C52 (0.002)2  C53 (0.002)3  ]
 243[1  5(0.002)  10(0.000004)  10(0.000000008)  ]
 243(1  0.01  0.00004  0.00000008  )
 243  2.43  0.00972  0.00001944 
 245.440 (correct to 3 decimal places)

Example 9

By substituting 0.001 for x in the expansion of (1 – x)7, find the value of (1.998)7 correct to
six significant figures.

Solution:

(1.998)7  (2  0.002)7
 27 (1  0.001)7
 27 [17  C17 (0.001)  C72 (0.001) 2  C37 (0.001)3  ]
 128[1  7(0.001)  21(0.000001)  35(0.000000001)  ]
 128(1  0.007  0.000021  0.000000035  )
 128  0.896  0.002688  0.00000448 
 127.107 (correct to 6 significant figures)

2-10