The Binomial Theorem

By studying the expanded form of each binomial expression, we are able to discover the following patterns in the resulting polynomials.

1. The first term is an. The exponent on a decreases by 1 in each successive term.

2. The exponents on b increase by 1 in each successive term. In the first term, the exponent on b is 0. (Because b0 1, b is not shown in the first term.) The last term is bn.

3. The sum of the exponents on the variables in any term is equal to n, the exponent on (a b)n.

4. There is one more term in the polynomial expansion than there is in the power of the binomial, n. There are n 1 terms in the expanded form of

(a b)n.

Using these observations, the variable parts of the expansion (a b)6 are

a6, a5b, a4b2, a3b3, a2b4, ab5, b6.

Patterns in Binomial Expansions

Let's now establish a pattern for the coefficients of the terms in the binomial expansion. Notice that each row in the figure begins and ends with 1. Any other number in the row can be obtained by adding the two numbers immediately above it.

Coefficients for (a b)1.

Coefficients for (a b)2.

Coefficients for (a b)3.

Coefficients for (a b)4.

Coefficients for (a b)5.

Coefficients for (a b)6.

• 1• 2 1

• 3 3 1• 4 6 4 1

• 5 10 10 5 11 6 15 20 15 6 1

The above triangular array of coefficients is called Pascal’s triangle. We can use the numbers in the sixth row and the variable parts we found to write the expansion for (a b)6. It is

(a b)6 a6 6a5b 15a4b2 20a3b3 15a2b4 6ab5 b6

Patterns in Binomial Expansions

Definition of a Binomial

Coefficient .For nonnegative integers n and r, with

n > r, the expression is called a

binomial coefficient and is defined by

r

n

r

n

)!(!

!

rnr

n

r

n

Example

3

7• Evaluate

Solution:

351*2*3*4*1*2*3

1*2*3*4*5*6*7

!4!3

!7

)!37(!3

!7

3

7

)!(!

!

rnr

n

r

n

A Formula for Expanding Binomials: The Binomial Theorem

nnnnn bn

nba

nba

na

nba

...

310)( 221

• For any positive integer n,

Example3)4( x• Expand

Solution:

3223

3

43

34*

2

34*

1

3

0

3

)4(

xxx

x

Example cont.3)4( x• Expand

Solution:

644812

64!0!3

!316

!1!2

!34

!2!1

!3

!3!0

!3

43

34*

2

34*

1

3

0

3

)4(

23

23

3223

3

xxx

xxx

xxx

x

Finding a Particular Term in a Binomial Expansion

The rth term of the expansion of (a+b)n is

11

1

rrn bar

n

ExampleFind the third term in the expansion of (4x-2y)8

(4x-2y)8 n=8, r=3, a=4x, b=-2ySolution:

2626

26

13138

11

752,4584)4096(28

)2()4(!6!2

!8

)2()4(13

8

1

yxyx

yx

yx

bar

n rrn

Find the fourth term in the expansion of (3x 2y)7.

Text Example

7

3

(3x)7 3(2y)3

7

3

(3x)4 (2y)3

7!

3!(7 3)!(3x)4 (2y)3

Solution We will use the formula for the rth term of the expansion (a b)n,

to find the fourth term of (3x 2y)7. For the fourth term of (3x 2y)7, n 7, r 4, a 3x, and b 2y. Thus, the fourth term is

7!

3!4!(81x4 )(8y3 )

7 6 54!

32 14!(81x 4 )(8y3) 35(81x 4)(8y3) 22,680x 4y3

The Binomial Theorem