Applications of binomial theorem

1. The Binomial Theorem
Digital Lesson

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The binomial theorem provides a useful method for raising any
binomial to a nonnegative integral power.
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Notice that each expansion has n + 1 terms.
1 term
2 terms
3 terms
4 terms
5 terms
6 terms
Example: (x + y)10 will have 10 + 1, or 11 terms.

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Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
1. The exponents on x decrease from n to 0.
The exponents on y increase from 0 to n.
2. Each term is of degree n.
Example: The 5th term of (x + y)10 is a term with x6y4.”

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The coefficients of the binomial expansion are called binomial
coefficients. The coefficients have symmetry.
The coefficient of xn–ryr in the expansion of (x + y)n is written
or nCr .
n
r
 
 
 
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
The first and last coefficients are 1.
The coefficients of the second and second to last terms
are equal to n.
1 1
Example: What are the last 2 terms of (x + y)10 ? Since n = 10,
the last two terms are 10xy9 + 1y10.
So, the last two terms of (x + y)10 can be expressed
as 10C9 xy9 + 10C10 y10 or as xy9 + y10.








9
10








10
10

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The triangular arrangement of numbers below is called Pascal’s
Triangle.
Each number in the interior of the triangle is the sum of the two
numbers immediately above it.
The numbers in the nth row of Pascal’s Triangle are the binomial
coefficients for (x + y)n .
1 1 1st row
1 2 1 2nd row
1 3 3 1 3rd row
1 4 6 4 1 4th row
1 5 10 10 5 1 5th row
0th row
1
6 + 4 = 10
1 + 2 = 3

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Example: Use the fifth row of Pascal’s Triangle to generate the
sixth row and find the binomial coefficients , , 6C4 and 6C2 .
6
1
 
 
 
6
5
 
 
 
5th row 1 5 10 10 5 1
6th row
6
0
 
 
 
6
1
 
 
 
6
2
 
 
 
6
3
 
 
 
6
4
 
 
 
6
5
 
 
 
6
6
 
 
 
6C0 6C1 6C2 6C3 6C4 6C5 6C6
= 6 = and 6C4 = 15 = 6C2.
6
1
 
 
 
6
5
 
 
 
There is symmetry between binomial coefficients.
nCr = nCn–r
1 6 15 20 15 6 1

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Example: Use Pascal’s Triangle to expand (2a + b)4.
(2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4
= 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4
= 16a4 + 32a3b + 24a2b2 + 8ab3 + b4
1 1 1st row
1 2 1 2nd row
1 3 3 1 3rd row
1 4 6 4 1 4th row
0th row
1

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The symbol n! (n factorial) denotes the product of the first n
positive integers. 0! is defined to be 1.
n! = n(n – 1)(n – 2)  3 • 2 • 1
1! = 1
4! = 4 • 3 • 2 • 1 = 24
6! = 6 • 5 • 4 • 3 • 2 • 1 = 720
Formula for Binomial Coefficients For all nonnegative
integers n and r, !
( )! !
n r
n
C
n r r


Example:
)
1
2
3
(
)
1
2
3
4
(
)
1
2
3
(
)
4
5
6
7
(
•
•
•
•
•
•
•
•
•
•
•
•

!
3
!
4
7
!
3
!
4
!
7
!
3
)!
3
7
(
!
7
•
•
•
3
7 



C
35
1
2
3
4
4
5
6
7
•
•
•
•
•
•



9. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9
Example: Use the formula to calculate the binomial coefficients
10C5, 15C0, and .
50
48
 
 
 
12
1
 
 
 
!
5
)!
5
10
(
!
10
•
5
10


C
!
0
)!
0
10
(
!
10
•
0
10


C
!
48
)!
48
50
(
!
50
48
50
•










!
1
)!
1
12
(
!
12
1
12
•










!
5
!
5
!
10
•

!
5
!
5
!
5
)
6
7
8
9
10
(
•
•
•
•
•
•
 252
1
2
3
4
5
6
7
8
9
10
•
•
•
•
•
•
•
•


!
0
!
10
!
10
•
 1
1
1
!
0
!
1



!
48
!
2
!
48
)
49
50
(
!
48
!
2
!
50
•
•
•
•

 1225
1
2
49
50
•
•


!
1
!
11
!
11
12
!
1
!
1
!
12
•
•
•

 12
1
12



10. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10
Binomial Theorem
1 1
( )n n n n r r n n
n r
x y x nx y C x y nxy y
  
       
!
with
( )! !
n r
n
C
n r r


Example: Use the Binomial Theorem to expand (x4 + 2)3.
0
3C 1
3C 2
3C 3
3C

 3
4
)
2
(x 
3
4
)
(x 
)
2
(
)
( 2
4
x 
2
4
)
2
)(
(x 3
)
2
(
1 
3
4
)
(x
 3 
)
2
(
)
( 2
4
x 3 
2
4
)
2
)(
(x 1
3
)
2
(
8
12
6 4
8
12



 x
x
x

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Although the Binomial Theorem is stated for a binomial which
is a sum of terms, it can also be used to expand a difference of
terms.
Simply rewrite
(x + y)n as (x + (– y))n
and apply the theorem to this sum.
Example: Use the Binomial Theorem to expand (3x – 4)4.
4
)
4
3
( 
x 4
))
4
(
3
( 

 x
4
3
2
2
3
4
)
4
(
1
)
4
)(
3
(
4
)
4
(
)
3
(
6
)
4
(
)
3
(
4
)
3
(
1 







 x
x
x
x
256
)
64
)(
3
(
4
)
16
)(
9
(
6
)
4
)(
27
(
4
81 2
3
4






 x
x
x
x
256
768
864
432
81 2
3
4




 x
x
x
x

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Example: Use the Binomial Theorem to write the first three
terms in the expansion of (2a + b)12 .
...
)
2
(
2
12
)
2
(
1
12
)
2
(
0
12
)
2
( 2
10
11
12
12




























 b
a
b
a
a
b
a
...
)
2
(
!
2
)!
2
12
(
!
12
)
2
(
!
1
)!
1
12
(
!
12
)
2
(
!
0
)!
0
12
(
!
12 2
10
10
11
11
12
12
•
•
•






 b
a
b
a
a
...
)
2
)(
11
12
(
)
2
(
12
)
2
( 2
10
10
11
11
12
12 • 


 b
a
b
a
a
...
135168
24576
4096 2
10
11
12



 b
a
b
a
a

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Example: Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y0 . The power of
y is 1 less than the number of the term in the expansion.
The eighth term is 13C7 x6y7.
Therefore, the eighth term of (x + y)13 is 1716 x6y7.
!
7
!
6
!
7
)
8
9
10
11
12
13
(
!
7
!
6
!
13
•
•
•
•
•
•
•
•
7
13 

C
1716
1
2
3
4
5
6
8
9
10
11
12
13
•
•
•
•
•
•
•
•
•
•

