Binomial Expansion for Positive Integer n — Edexcel A-level Mathematics (Pure)

SPEC 4.1 • SEQUENCES AND SERIES

Binomial Expansion for Rational n

Learn to expand expressions of the form (1 + x)^n where n is negative or fractional, and use expansions for approximations

Learning Objectives

By the end of this lesson, you will be able to:

Apply the generalised binomial formula for rational values of n

Determine the validity condition for convergence

Expand expressions of the form

(1 + x)^{n}

where n is rational

Manipulate expressions into the required form

(1 + something)^{n}

Find the validity of combined binomial expressions

Use binomial expansions to find approximations

Recap: Positive Integer Powers

In the previous lesson, we expanded

(a + b)^{n}

where n was a positive integer.

Formula

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

Key Features for Positive Integer n

The expansion has exactly

n + 1

terms

The expansion is finite (it terminates)

It works for any values of a and b

But what happens when n is negative or a fraction?

The Problem with Factorials

Recall that

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

Problem

Factorials are only defined for non-negative integers!

We cannot calculate

(- 2)!

or

(\frac{1}{2})!

using the factorial formula.

So we need a different approach for rational values of n.

Solution

Instead of using

(r n)

, we write out the multiplication pattern directly:

\frac{n ( n - 1 ) ( n - 2 ) ... ( n - r + 1 )}{r !}

This works for ANY value of n - positive, negative, or fractional!

The Generalised Binomial Formula

Formula

For rational n and

∣ x ∣ < 1

:

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + ...

Critical: Validity Condition

This expansion is only valid when

∣ x ∣ < 1

, i.e.,

- 1 < x < 1

Unlike positive integer powers, this expansion goes on forever (infinite series)!

Remember

The expansion must start with

(1 + something)^{n}

- not

(a + b)^{n}

. We'll see how to convert other forms shortly.

Understanding the Pattern

Let's break down each term in the expansion:

TermCoefficientFull Term

1st1

1

2nd

n

n x

3rd

\frac{n ( n - 1 )}{2 !}

\frac{n ( n - 1 )}{2 !} x^{2}

4th

\frac{n ( n - 1 ) ( n - 2 )}{3 !}

\frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3}

5th

\frac{n ( n - 1 ) ( n - 2 ) ( n - 3 )}{4 !}

\frac{n ( n - 1 ) ( n - 2 ) ( n - 3 )}{4 !} x^{4}

Pattern

Each numerator has one more factor than the previous, decreasing from n. The denominator is always r! for the (r+1)th term.

Worked Example01

Expand $(1 + x)^{- 1}$ up to and including the term in $x^{3}$ .

Here

n = - 1

. Substituting into the formula:

1

1st term:

1

2

2nd term:

n x = (- 1) x = - x

3

3rd term:

\frac{( - 1 ) ( - 2 )}{2 !} x^{2} = \frac{2}{2} x^{2} = x^{2}

4

4th term:

\frac{( - 1 ) ( - 2 ) ( - 3 )}{3 !} x^{3} = \frac{- 6}{6} x^{3} = - x^{3}

Answer

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

Valid for

∣ x ∣ < 1

Worked Example02

Expand $\frac{1}{1 + x}$ up to and including the term in $x^{3}$ .

First, rewrite:

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

. Here

n = - \frac{1}{2}

.

1

1st term:

1

2

2nd term:

- \frac{1}{2} x

3

3rd term:

\frac{( - \frac{1}{2} ) ( - \frac{3}{2} )}{2 !} x^{2} = \frac{\frac{3}{4}}{2} x^{2} = \frac{3}{8} x^{2}

4

4th term:

\frac{( - \frac{1}{2} ) ( - \frac{3}{2} ) ( - \frac{5}{2} )}{3 !} x^{3} = \frac{- \frac{15}{8}}{6} x^{3} = - \frac{5}{16} x^{3}

Answer

(1 + x)^{- \frac{1}{2}} = 1 - \frac{1}{2} x + \frac{3}{8} x^{2} - \frac{5}{16} x^{3} + ...

Valid for

∣ x ∣ < 1

Converting to the Required Form

Important

The formula only works for

(1 + something)^{n}

. If your expression looks different, you must convert it first!

Factor Out a Constant

(a + b x)^{n} = a^{n} (1 + \frac{b x}{a})^{n}

Example 1:

(4 + x)^{- 1} = 4^{- 1} (1 + \frac{x}{4})^{- 1} = \frac{1}{4} (1 + \frac{x}{4})^{- 1}

Example 2:

(9 - 3 x)^{\frac{1}{2}} = 9^{\frac{1}{2}} (1 - \frac{x}{3})^{\frac{1}{2}} = 3 (1 - \frac{x}{3})^{\frac{1}{2}}

Don't Forget

When you factor out

a

, you must raise it to the power n as well!

Worked Example03

Expand $(4 + x)^{- \frac{1}{2}}$ up to and including the term in $x^{2}$ . State the values of x for which the expansion is valid.

Step 1: Convert to standard form

(4 + x)^{- \frac{1}{2}} = 4^{- \frac{1}{2}} (1 + \frac{x}{4})^{- \frac{1}{2}} = \frac{1}{2} (1 + \frac{x}{4})^{- \frac{1}{2}}

Step 2: Expand

(1 + \frac{x}{4})^{- \frac{1}{2}}

using

n = - \frac{1}{2}

:

= 1 - \frac{1}{2} \cdot \frac{x}{4} + \frac{3}{8} \cdot \frac{x ^{2}}{16} + ... = 1 - \frac{x}{8} + \frac{3 x ^{2}}{128} + ...

Step 3: Multiply by

\frac{1}{2}

:

Validity:

\frac{x}{4} < 1

, so

∣ x ∣ < 4

Answer

(4 + x)^{- \frac{1}{2}} = \frac{1}{2} - \frac{x}{16} + \frac{3 x ^{2}}{256} + ...

Determining Validity

The expansion is valid when the "x part" inside the bracket has absolute value less than 1.

Method

Identify what replaces x in

(1 + x)^{n}

Set

∣ that expression ∣ < 1

Solve the inequality

ExpressionThe "x part"Validity Condition

(1 + 3 x)^{- 2}

3 x

∣3 x ∣ < 1 \Rightarrow ∣ x ∣ < \frac{1}{3}

(1 - \frac{x}{2})^{\frac{1}{2}}

- \frac{x}{2}

- \frac{x}{2} < 1 \Rightarrow ∣ x ∣ < 2

(4 + x)^{- 1} = \frac{1}{4} (1 + \frac{x}{4})^{- 1}

\frac{x}{4}

\frac{x}{4} < 1 \Rightarrow ∣ x ∣ < 4

Validity of Combined Expressions

When adding or subtracting two binomial expansions, you must find where both are valid.

Method

Find the validity of each expansion separately

Take the intersection (overlap) of both conditions

The combined expansion is valid in this intersection

Example

If expansion A is valid for

∣ x ∣ < 2

and expansion B is valid for

∣ x ∣ < \frac{1}{3}

:

The combined expression A + B is valid for

∣ x ∣ < \frac{1}{3}

(We take the more restrictive condition)

Worked Example04

Expand $\frac{1}{1 + x} + \frac{1}{1 + 3 x}$ up to and including the term in $x^{2}$ . State the values of x for which the expansion is valid.

Expand $(1 + x)^{- 1}$ :

= 1 - x + x^{2} + ...

Valid for

∣ x ∣ < 1

Expand $(1 + 3 x)^{- 1}$ :

= 1 + (- 1) (3 x) + \frac{( - 1 ) ( - 2 )}{2 !} (3 x)^{2} + ... = 1 - 3 x + 9 x^{2} + ...

Valid for

∣3 x ∣ < 1

, i.e.,

∣ x ∣ < \frac{1}{3}

Add:

(1 - x + x^{2}) + (1 - 3 x + 9 x^{2}) = 2 - 4 x + 10 x^{2} + ...

Validity: Intersection of

∣ x ∣ < 1

and

∣ x ∣ < \frac{1}{3}

is

∣ x ∣ < \frac{1}{3}

Answer

\frac{1}{1 + x} + \frac{1}{1 + 3 x} = 2 - 4 x + 10 x^{2} + ...

Using Expansions for Approximations

Binomial expansions give us polynomial approximations for complicated expressions. When x is small, we can use just the first few terms.

Key Idea

If

∣ x ∣

is small, higher powers of x become negligible:

x = 0.01 \Rightarrow x^{2} = 0.0001 \Rightarrow x^{3} = 0.000001

So truncating the series gives a good approximation!

Method for Approximations

Write the expression in the form

(1 + x)^{n}

or

a^{n} (1 + \frac{b x}{a})^{n}

Expand to the required number of terms

Check that your x value satisfies the validity condition

Substitute the value and calculate

Choosing Valid x Values

When using approximations, you must ensure your substitution satisfies the validity condition.

Example: Why

3

Cannot Use

(1 + x)^{\frac{1}{2}}

Directly

To approximate

3

, you might try

3 = 1 + 2 = (1 + 2)^{\frac{1}{2}}

But this requires

x = 2

, and we need

∣ x ∣ < 1

for validity!

This substitution is INVALID — the expansion won't converge.

Solution: Rewrite to Get a Small x

Instead, write

3 = 4 - 1 = 4 (1 - \frac{1}{4}) = 2 (1 - \frac{1}{4})^{\frac{1}{2}}

Now

x = - \frac{1}{4}

, and

∣ - \frac{1}{4} ∣ = 0.25 < 1

✓

This substitution IS VALID!

Strategy

Find a nearby perfect power (square, cube, etc.) and express your number relative to it.

Worked Example05

Use the binomial expansion to find an approximation for $1.02$ correct to 5 decimal places.

Step 1: Write in binomial form

1.02 = (1 + 0.02)^{\frac{1}{2}} = (1 + x)^{\frac{1}{2}}

where

x = 0.02

Step 2: Expand

(1 + x)^{\frac{1}{2}}

= 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} + ...

Step 3: Substitute

x = 0.02

= 1 + \frac{1}{2} (0.02) - \frac{1}{8} (0.0004) + ...

= 1 + 0.01 - 0.00005 + ... \approx 1.00995

Answer

1.02 \approx 1.00995

(Calculator: 1.009950494...)

Worked Example06

Use the binomial expansion to find an approximation for $\frac{1}{9.1}$ correct to 4 decimal places.

Step 1: Convert to binomial form

\frac{1}{9.1} = (9.1)^{- \frac{1}{2}} = (9 + 0.1)^{- \frac{1}{2}} = 9^{- \frac{1}{2}} (1 + \frac{1}{90})^{- \frac{1}{2}} = \frac{1}{3} (1 + \frac{1}{90})^{- \frac{1}{2}}

Step 2: Expand with

x = \frac{1}{90}

(1 + x)^{- \frac{1}{2}} = 1 - \frac{1}{2} x + \frac{3}{8} x^{2} + ...

= 1 - \frac{1}{180} + \frac{3}{64800} + ... \approx 0.99444

Step 3: Multiply by

\frac{1}{3}

Answer

\frac{1}{9.1} \approx \frac{0.99444}{3} \approx 0.3315

(Calculator: 0.331479...)

Practice

Question 1: Expand

(1 + x)^{- 3}

up to and including the term in

x^{3}

. State the values of x for which the expansion is valid.

Using

n = - 3

:

1st:

1

, 2nd:

- 3 x

, 3rd:

\frac{( - 3 ) ( - 4 )}{2 !} x^{2} = 6 x^{2}

, 4th:

\frac{( - 3 ) ( - 4 ) ( - 5 )}{3 !} x^{3} = - 10 x^{3}

Answer

(1 + x)^{- 3} = 1 - 3 x + 6 x^{2} - 10 x^{3} + ...

Valid for

∣ x ∣ < 1

Practice

Question 2: Expand

\frac{1}{( 1 + 2 x ) ^{2}} + \frac{1}{1 - x}

up to and including the term in

x^{2}

. State the validity.

(1 + 2 x)^{- 2} = 1 - 4 x + 12 x^{2} + ...

Valid for

∣ x ∣ < \frac{1}{2}

(1 - x)^{- 1} = 1 + x + x^{2} + ...

Valid for

∣ x ∣ < 1

Sum:

(1 - 4 x + 12 x^{2}) + (1 + x + x^{2}) = 2 - 3 x + 13 x^{2} + ...

Answer

2 - 3 x + 13 x^{2} + ...

Valid for

∣ x ∣ < \frac{1}{2}

(more restrictive)

Practice

Question 3: Use the binomial expansion to find an approximation for

3 1.03

correct to 5 decimal places.

3 1.03 = (1 + 0.03)^{\frac{1}{3}}

Expand:

(1 + x)^{\frac{1}{3}} = 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + ...

Substitute

x = 0.03

:

= 1 + \frac{1}{3} (0.03) - \frac{1}{9} (0.0009) + ... = 1 + 0.01 - 0.0001 + ... \approx 1.0099

Answer

3 1.03 \approx 1.00990

(Calculator: 1.009901635...)

Summary

Key Formulas

NameFormula

Generalised Binomial

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + ...

Converting Form

(a + b x)^{n} = a^{n} (1 + \frac{b x}{a})^{n}

Validity

\frac{b x}{a} < 1 \Rightarrow ∣ x ∣ < \frac{∣ a ∣}{∣ b ∣}

Key Points

Expansion is infinite (never terminates) for non-positive-integer n

Always state the validity condition

For combined expressions, take the more restrictive validity

For approximations, substitute small values of x