What is not factoring out the constant?

(4 + x)^1/2 must become 2(1 + (x)/(4))^1/2 before expanding; expanding (4 + x) directly is wrong.

What are sign errors with negative indices?

The coefficients n(n-1) alternate in sign for negative n; compute them carefully term by term.

SEAB Original-style practice: By writing sqrt(4 + x) = 2(1 + x4)^1/2, expand sqrt(4 + x) in ascending powers of x up to the term in x^2, and state the range of validity.

sqrt(4 + x) = 2(1 + x4)^1/2. Expand with n = 12 and variable x4: (1 + x4)^1/2 = 1 + 12(x4) + (1)/(2)(-(1)/(2))2!(x4)^2 + = 1 + x8 - x^2128 + Multiply by 2: sqrt(4 + x) = 2 + x4 - x^264 + Valid when |x4| < 1, that is |x| < 4. Markers reward taking the factor of 2 out first, the correct coefficients, and the validity |x| < 4.

SingaporeMathsSyllabus dot point

How do we expand a binomial raised to a rational power, and when is the expansion valid?

Expand (1 + x) to the power n for rational n as a series, state the range of validity, and use the expansion to obtain approximations

A focused answer to the H2 Mathematics outcome on the binomial expansion for rational index. The general series, the range of validity, handling expressions not in standard form, and using the expansion for approximations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to expand $(1 + x)^n$ for rational $n$ (including negative and fractional indices) as an infinite series, state the range of $x$ for which it is valid, manipulate non-standard expressions into the standard form, and use truncated expansions to obtain numerical approximations.

The answer

The general expansion

For any rational $n$ and $|x| < 1$ :

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

Unlike the positive-integer case, this series does not terminate; it continues indefinitely and only equals $(1+x)^n$ within its range of validity.

Range of validity

The expansion of $(1 + x)^n$ converges only for $|x| < 1$ . When the variable is some multiple $kx$ , the condition becomes $|kx| < 1$ , that is $|x| < \dfrac{1}{|k|}$ . Always state this.

Getting to standard form

Many expressions are not $(1 + x)^n$ directly. Factor out the leading constant so the bracket starts with $1$ :

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

Then expand the bracket and multiply through by $a^n$ . The validity becomes $\left|\dfrac{b}{a}x\right| < 1$ .

Using the expansion to approximate

Substituting a small numerical value of $x$ into the first few terms gives a good approximation, with smaller $x$ giving faster convergence. This is how the expansion produces decimal estimates of roots and reciprocals.

Worked example

Expand $\dfrac{1}{\sqrt{1 - 3x}}$ in ascending powers of $x$ up to the term in $x^2$ , state the validity, and use it to estimate $\dfrac{1}{\sqrt{0.97}}$ .

Step 1: Write in index form

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

Step 2: Apply the binomial series with n = -1/2 and variable -3x

1 + \left(-\tfrac{1}{2}\right)(-3x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-3x)^2 + \cdots

Step 3: Simplify each term

First term $1$ . Second: $\tfrac{3}{2}x$ . Third: $\dfrac{\frac{3}{4}}{2}(9x^2) = \dfrac{3}{8}\times 9 x^2 = \dfrac{27}{8}x^2$ .

\frac{1}{\sqrt{1 - 3x}} = 1 + \frac{3}{2}x + \frac{27}{8}x^2 + \cdots

Step 4: State the validity

Valid when $|3x| < 1$ , that is $|x| < \dfrac{1}{3}$ .

Step 5: Estimate

$\dfrac{1}{\sqrt{0.97}} = \dfrac{1}{\sqrt{1 - 0.03}}$ , so $x = 0.01$ . Then $\approx 1 + \dfrac{3}{2}(0.01) + \dfrac{27}{8}(0.0001) = 1 + 0.015 + 0.0003375 \approx 1.0153$ .

Examples in context

Example 1. Quick reciprocal estimate. $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \cdots$ for $|x| < 1$ recovers the geometric series, so $\dfrac{1}{1.02} \approx 1 - 0.02 + 0.0004 = 0.9804$ , matching the true value to four decimal places.

Example 2. Relativity-style approximation. A factor $\left(1 - \dfrac{v^2}{c^2}\right)^{-1/2} \approx 1 + \dfrac{1}{2}\dfrac{v^2}{c^2}$ for small $v/c$ uses the rational-index expansion to linearise a square-root expression, the standard low-speed approximation.

Try this

Q1. Expand $(1 - x)^{1/2}$ up to the term in $x^2$ . [3 marks]

Cue. $1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \cdots$ , valid for $|x| < 1$ .

Q2. State the range of validity of the expansion of $(1 + 4x)^{-1}$ . [1 mark]

Cue. $|4x| < 1$ , so $|x| < \dfrac{1}{4}$ .

Q3. Write $(9 + x)^{1/2}$ in a form ready for the binomial expansion. [2 marks]

Cue. $3\left(1 + \dfrac{x}{9}\right)^{1/2}$ , valid for $|x| < 9$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksExpand

(1 + 2x)^{-2}

in ascending powers of

x

up to and including the term in

x^3

, and state the range of values of

x

for which the expansion is valid.

Show worked answer →

With $n = -2$ and the variable $2x$ :
$(1 + 2x)^{-2} = 1 + (-2)(2x) + \dfrac{(-2)(-3)}{2!}(2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(2x)^3 + \cdots$

$= 1 - 4x + 3(4x^2) - 4(8x^3) + \cdots = 1 - 4x + 12x^2 - 32x^3 + \cdots$

Valid when $|2x| < 1$ , that is $|x| < \dfrac{1}{2}$ .

Markers reward the binomial coefficients for negative index, substituting $2x$ correctly (including powers of $2$ ), the first four terms, and the validity condition $|x| < \frac{1}{2}$ .

Original5 marksBy writing

\sqrt{4 + x} = 2\left(1 + \dfrac{x}{4}\right)^{1/2}

, expand

\sqrt{4 + x}

in ascending powers of

x

up to the term in

x^2

, and state the range of validity.

Show worked answer →

$\sqrt{4 + x} = 2\left(1 + \dfrac{x}{4}\right)^{1/2}$ .

Expand with $n = \tfrac{1}{2}$ and variable $\dfrac{x}{4}$ :
$\left(1 + \dfrac{x}{4}\right)^{1/2} = 1 + \tfrac{1}{2}\left(\dfrac{x}{4}\right) + \dfrac{\frac{1}{2}(-\frac{1}{2})}{2!}\left(\dfrac{x}{4}\right)^2 + \cdots = 1 + \dfrac{x}{8} - \dfrac{x^2}{128} + \cdots$

Multiply by $2$ : $\sqrt{4 + x} = 2 + \dfrac{x}{4} - \dfrac{x^2}{64} + \cdots$

Valid when $\left|\dfrac{x}{4}\right| < 1$ , that is $|x| < 4$ .

Markers reward taking the factor of $2$ out first, the correct coefficients, and the validity $|x| < 4$ .