At the end of the lesson students will be able to:

a) expand where n is a positive

integer using Binomial theorem and

Pascal triangle.

LEARNING OUTCOME

nba )(

Binomial:

Polynomial with two terms

Example:

3x2 – 4, x+1 , a-b

Binomial expansion:

Expansion of binomial expression raised to a

power of n.

Example:

(x+2x2)2 = (x+2x2)(x+2x2) = x2 + 4x3 + 4x4

Consider the patterns formed by expanding (a + b)n.

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Notice that each expansion has n + 1 terms.

1 term

2 terms

3 terms

4 terms

5 terms

6 terms

Example: (a + b)10 will have 10 + 1, or 11 terms.

Consider the patterns formed by expanding (a + b)n.

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b +3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 +4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 +10a2b3 + 5ab4 +b5

1. The power on a decrease from n to 0.

The power on b increase from 0 to n.

2. Each term is of degree n.

Example: The 5th term of (a + b)10 is a term with a6b4.”

The powers of a is decrease by 1 in each successive term,

and the powers of b increase by 1 in each successive

term.

There are several patterns that all of the binomial

expansions have:

The number of terms in each resulting polynomial is

always one more than the power of the binomial. Thus,

there are n+1 terms in each expansion.

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4n=4:

5 terms

(a + b)4 = a4 b0 + 4a3 b1 + 6a2 b2 +4a1b3 + a0b4

1.

2.

3.The sum of the powers of each term is n

(a + b)4 = a4b0 + 4a3b1 + 6a2b2 +4a1b3 + a0b4

4. The coefficients increase and decrease in a symmetric

manner

(a + b)4 = a4b0 + 4a3b1 + 6a2b2 +4a1b3 + a0b41 1

The first and last coefficients are 1.

The coefficients of the second and second to last terms

are equal to n.

4+0=4 1+3=4

Using this pattern, we can develop a

generalized formula for (a + b)n

(a + b)n = an b0 + an-1b1 + an-2b2

+…+ a1bn-1 + a0bn

What coefficients go in the

blanks?

The binomial coefficients are the coefficients in

the expansion of . Let n and r be whole

numbers with , then

DEFINITION OF BINOMIAL COEFFICIENT

rn

!)!(

!

rrn

nC

r

nr

n

nba

a)

b)

10!2)!25(

!5

2

52

5

C

84!3)!39(

!9

3

93

9

C

EXAMPLE

Let’s return to the question of the binomial

expansion and how to determine the

coefficients:

(a + b)n = an b0 + an-1b1+ an-2b2

+…+ a1bn-1 + a0bn

1n

n

1

n

0

n

2

n

n

n

The Binomial Theorem

Binomial Theorem

Let a and b be a real numbers, then for any

positive integers n,

nrrn

nnnn

ban

nba

r

n

ban

ban

ban

ba

0

22110

...

...210

)(

OR

rrnn

r

rn

n baCba

0

)(

Expand using Binomial theorem 33x

SOLUTION

rrnn

r

rn

n baCba

0

)(

rr

r

r xCx 3)3( 33

0

3

3

Subst.

a= x, b=3 ,

and n=3

EXAMPLE

Expand the summation

3)3(x 033

0

3x

12

31

3x

213

2

3x

30

33

3x

279333 223 xxx

27279 223 xxx

These numbers

will always be

the same.

a)5)2( x

EXAMPLE

4)21( xb)

PASCAL’S

TRIANGLE

Consider the patterns formed by expanding (a + b)n.

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

3)( ba

2)( ba

1)( ba

0)( ba

4)( ba

Expression Coefficients

0

0

0

1

1

1

1

3

2

2

1

2

0

2

0

4

3

3

2

3

0

3

4

4

3

4

1

4

2

4

1 2 1

1 3 3 1

1 1

1 4 6 4 1

1

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 (a + b)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 row1

6 + 4 = 10

1 + 2 = 33)( ba

2)( ba

1)( ba

0)( ba

4)( ba

Expression Coefficients

5)( ba

These triangular arrangement of numbers is called Pascal’s

Triangle.

Find the coefficients in the expansion of 6)( ba

We need 7 rows

1 2 1

1 3 3 1

1 1

1

1 4 6 4 1

1 5 10 110 5

1 6 15 120 15 6Coefficients

Solution

EXAMPLE

Pascal’s triangle gives the coefficients

1 6 15 120 15 6

The full expansion is

6051423324506 1615201561 bababababababa

Tip: The powers in each term sum to 6

Answer:

EXERCISE

Use Binomial theorem to expand the following

binomial expression.

642

246 161520156

xxxxxx

i) 4)3

1(a

ii)

432

8110854121

aaaa

6)1

(x

x

Answer:

EXERCISE

Use Pascal’s triangle to expand the following

binomial expression.

i) 3)31( y

4)51( xiii)

32 272791 yyy

432 625500150201 xxxx