Binomial Expansion for Positive Integer n

Learn to expand expressions of the form (a + b)^n using Pascal's triangle and binomial coefficients

SPEC 4.1 • SEQUENCES AND SERIES

What is a Binomial?

A binomial is an algebraic expression containing exactly two terms, such as:

(x + y)

(2 a - 3 b)

(1 + 3 x)

Key Point

The binomial expansion allows us to expand expressions of the form

(a + b)^{n}

without having to multiply out repeatedly.

You may already know these expansions:

(a + b)^{1} = a + b

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

But what about

(a + b)^{10}

? We need a systematic method!

Pascal's Triangle

Pascal's triangle gives us the coefficients for binomial expansions:

121 1331

14641

15101051

How It Works

Each number is the sum of the two numbers directly above it. Row n gives the coefficients for

(a + b)^{n}

.

Worked Example01

Use Pascal's triangle to expand $(x + y)^{4}$ .

Row 4 of Pascal's triangle gives us: 1, 4, 6, 4, 1

These are our coefficients. Now we apply the pattern:

1

(x + y)^{4} = 1 \cdot x^{4} + 4 \cdot x^{3} y + 6 \cdot x^{2} y^{2} + 4 \cdot x y^{3} + 1 \cdot y^{4}

2

(x + y)^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}

Factorial Notation

Definition

n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1

Examples:

5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

4! = 4 \times 3 \times 2 \times 1 = 24

3! = 3 \times 2 \times 1 = 6

Special Cases

$0! = 1$ (by definition)

$1! = 1$

Factorials grow very quickly:

10! = 3, 628, 800

Binomial Coefficients

Formula

(r n) = \frac{n !}{r ! ( n - r )!}

This is read as "n choose r" and can also be written as

^{n} C_{r}

or

C (n, r)

.

PropertyFormula

First coefficient

(0 n) = 1

Last coefficient

(n n) = 1

Symmetry

(r n) = (n - r n)

Using Your Calculator

Your Casio scientific calculator has a built-in function for calculating binomial coefficients

(r n)

or

^{n} C_{r}

.

Casio fx-83GT / fx-85GT / fx-991 Series

To calculate $(3 8)$ (or $^{8} C_{3}$ ):

1

Type 8

2

Press SHIFT then × (this gives you nCr)

3

Type 3

4

Press =

✓

Answer: 56

Tip

On newer Casio ClassWiz models (fx-991EX), press OPTN → Probability → nCr instead.

Worked Example02

Calculate $(3 8)$ using the formula and verify with your calculator.

Using

(r n) = \frac{n !}{r ! ( n - r )!}

with

n = 8

and

r = 3

:

1

(3 8) = \frac{8 !}{3 ! ( 8 - 3 )!} = \frac{8 !}{3 ! \cdot 5 !}

2

= \frac{8 \times 7 \times 6 \times 5 !}{3 ! \times 5 !}

3

= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

Calculator check: 8 SHIFT × 3 = gives 56 ✓

The Binomial Theorem

(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}

In expanded form:

(a + b)^{n} = (0 n) a^{n} + (1 n) a^{n - 1} b + (2 n) a^{n - 2} b^{2} + \dots + (n n) b^{n}

Key Observations

The expansion has

(n + 1)

terms

Powers of

a

decrease from

n

to

0

Powers of

b

increase from

0

to

n

In each term, the powers of

a

and

b

sum to

n

Worked Example03

Expand $(2 x + 3)^{4}$ completely.

Using the binomial theorem with

a = 2 x

,

b = 3

,

n = 4

:

1

(2 x + 3)^{4} = (0 4) (2 x)^{4} + (1 4) (2 x)^{3} (3) + (2 4) (2 x)^{2} (3)^{2} + (3 4) (2 x) (3)^{3} + (4 4) (3)^{4}

2

= 1 (16 x^{4}) + 4 (8 x^{3}) (3) + 6 (4 x^{2}) (9) + 4 (2 x) (27) + 1 (81)

3

= 16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81

Finding Specific Terms

General Term Formula

T_{r + 1} = (r n) a^{n - r} b^{r}

The

(r + 1)

th term in the expansion of

(a + b)^{n}

is given by this formula.

Useful For

Finding the coefficient of a specific power of

x

Finding the term independent of

x

(constant term)

Finding the middle term(s)

Note: When n is even, there is one middle term. When n is odd, there are two middle terms.

Worked Example04

Find the coefficient of $x^{3}$ in the expansion of $(1 + 2 x)^{7}$ .

Here

a = 1

,

b = 2 x

,

n = 7

. We need the term containing

x^{3}

.

1

The general term is

T_{r + 1} = (r 7) (1)^{7 - r} (2 x)^{r} = (r 7) \cdot 2^{r} \cdot x^{r}

2

For

x^{3}

, we need

r = 3

3

T_{4} = (3 7) \cdot 2^{3} = 35 \times 8 = 280

Worked Example05

Find the term independent of $x$ in the expansion of $(x^{2} - \frac{1}{x})^{6}$ .

Here

a = x^{2}

,

b = - \frac{1}{x} = - x^{- 1}

,

n = 6

.

1

General term:

T_{r + 1} = (r 6) (x^{2})^{6 - r} (- \frac{1}{x})^{r}

2

= (r 6) (- 1)^{r} \cdot x^{2 (6 - r)} \cdot x^{- r} = (r 6) (- 1)^{r} \cdot x^{12 - 3 r}

3

For term independent of

x

:

12 - 3 r = 0

, so

r = 4

4

T_{5} = (4 6) (- 1)^{4} = 15 \times 1 = 15

Worked Example06

Find the first 4 terms of $(1 - 3 x)^{8}$ in ascending powers of $x$ .

Here

a = 1

,

b = - 3 x

,

n = 8

.

1

T_{1} = (0 8) (1)^{8} (- 3 x)^{0} = 1

2

T_{2} = (1 8) (1)^{7} (- 3 x)^{1} = 8 \times (- 3 x) = - 24 x

3

T_{3} = (2 8) (1)^{6} (- 3 x)^{2} = 28 \times 9 x^{2} = 252 x^{2}

4

T_{4} = (3 8) (1)^{5} (- 3 x)^{3} = 56 \times (- 27 x^{3}) = - 1512 x^{3}

Answer

1 - 24 x + 252 x^{2} - 1512 x^{3} + \dots

Worked Example07

The coefficient of $x^{2}$ in $(1 + a x)^{6}$ is 60. Find the value of $a$ .

The coefficient of

x^{2}

comes from the term where

r = 2

.

1

T_{3} = (2 6) (1)^{4} (a x)^{2} = 15 \cdot a^{2} \cdot x^{2}

2

Coefficient of

x^{2} = 15 a^{2}

3

Given:

15 a^{2} = 60

4

a^{2} = 4

, so

a = \pm 2

Answer

a = 2

or

a = - 2

Common Mistakes

Avoid These Errors

Forgetting $0! = 1$ — This is a defined value, not zero!

Miscounting terms —

(a + b)^{n}

has

n + 1

terms, not

n

Sign errors with negative $b$ — Remember

(- 3 x)^{3} = - 27 x^{3}

Forgetting to raise coefficients to powers — In

(2 x)^{3}

, both 2 and x are cubed:

8 x^{3}

Confusing powers of $a$ and $b$ — Power of

a

decreases, power of

b

increases

Practice

Question 1: Expand

(3 x - 2)^{4}

completely.

Using binomial theorem with

a = 3 x

,

b = - 2

,

n = 4

:

(3 x - 2)^{4} = (0 4) (3 x)^{4} + (1 4) (3 x)^{3} (- 2) + (2 4) (3 x)^{2} (- 2)^{2} + (3 4) (3 x) (- 2)^{3} + (4 4) (- 2)^{4}

= 81 x^{4} - 216 x^{3} + 216 x^{2} - 96 x + 16

Practice

Question 2: Find the coefficient of

x^{5}

in the expansion of

(2 + x)^{9}

.

General term:

T_{r + 1} = (r 9) (2)^{9 - r} (x)^{r}

For

x^{5}

:

r = 5

T_{6} = (5 9) (2)^{4} (x)^{5} = 126 \times 16 \times x^{5} = 2016 x^{5}

Coefficient = 2016

Practice

Question 3: Find the term containing

x^{4}

in the expansion of

(x + \frac{2}{x})^{8}

.

General term:

T_{r + 1} = (r 8) (x)^{8 - r} (\frac{2}{x})^{r} = (r 8) \cdot 2^{r} \cdot x^{8 - r} \cdot x^{- r} = (r 8) \cdot 2^{r} \cdot x^{8 - 2 r}

For

x^{4}

:

8 - 2 r = 4

, so

r = 2

T_{3} = (2 8) \cdot 2^{2} \cdot x^{4} = 28 \times 4 \times x^{4} = 112 x^{4}

Term = $112 x^{4}$

Summary

Key Formulas

NameFormula

Factorial

n! = n \times (n - 1) \times \dots \times 2 \times 1

Binomial Coefficient

(r n) = \frac{n !}{r ! ( n - r )!}

Binomial Theorem

(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}

General Term

T_{r + 1} = (r n) a^{n - r} b^{r}

Remember

(a + b)^{n}

has

n + 1

terms

0! = 1

Powers of

a

decrease while powers of

b

increase

Use your calculator: n SHIFT × r = for

^{n} C_{r}