Binomial Estimation

This lesson focuses on using the binomial expansion to approximate numerical values, determine accuracy, and establish ranges of validity. This is a common exam topic in Edexcel A-Level Mathematics (9MA0), often appearing as a multi-part question.

Recap: The General Binomial Expansion

For rational n and |x| < 1:

(1 + x)^n = 1 + nx + n(n-1)/2! x² + n(n-1)(n-2)/3! x³ + ...

For (a + bx)^n: factor out a^n first to get a^n(1 + bx/a)^n, valid for |bx/a| < 1.

Approximating Square Roots

Worked Example 1

Use a binomial expansion to find an approximation for sqrt(1.02), stating the expansion used.

sqrt(1.02) = (1 + 0.02)^(1/2)

Using (1 + x)^(1/2) ≈ 1 + (1/2)x - (1/8)x² + (1/16)x³

With x = 0.02:

≈ 1 + (1/2)(0.02) - (1/8)(0.0004) + (1/16)(0.000008)

= 1 + 0.01 - 0.00005 + 0.0000005

= 1.0099505

Calculator: sqrt(1.02) = 1.00995049...

The approximation is accurate to 6 decimal places.

Worked Example 2

Use a suitable binomial expansion to estimate sqrt(26).

sqrt(26) = sqrt(25 x (1 + 1/25)) = 5 x (1 + 0.04)^(1/2)

(1 + 0.04)^(1/2) ≈ 1 + (1/2)(0.04) - (1/8)(0.04)² + (1/16)(0.04)³

= 1 + 0.02 - 0.0002 + 0.000004

= 1.019804

So sqrt(26) ≈ 5 x 1.019804 = 5.09902

Calculator: sqrt(26) = 5.09901951...

Accurate to 4 decimal places.

Approximating Cube Roots

Worked Example 3

Estimate the cube root of 1030.

1030 = 1000(1 + 30/1000) = 1000(1 + 0.03)

Cube root of 1030 = 10 x (1 + 0.03)^(1/3)

(1 + x)^(1/3) ≈ 1 + (1/3)x + (1/3)(-2/3)/2! x² + (1/3)(-2/3)(-5/3)/3! x³

= 1 + x/3 - x²/9 + 5x³/81

With x = 0.03:

= 1 + 0.01 - 0.0001 + 0.0000015

= 1.0099015

Cube root of 1030 ≈ 10 x 1.0099015 = 10.099

Calculator: 1030^(1/3) = 10.09901...

Approximating Reciprocals

Worked Example 4

Use a binomial expansion to estimate 1/1.03.

1/1.03 = (1 + 0.03)^(-1)

(1 + x)^(-1) ≈ 1 - x + x² - x³ + ...

With x = 0.03:

≈ 1 - 0.03 + 0.0009 - 0.000027

= 0.970873

Calculator: 1/1.03 = 0.970874...

Worked Example 5

Estimate 1/(2.04)³.

1/(2.04)³ = (2 + 0.04)^(-3) = 2^(-3) x (1 + 0.02)^(-3)

= (1/8) x (1 + 0.02)^(-3)

(1 + x)^(-3) ≈ 1 + (-3)x + (-3)(-4)/2! x² + (-3)(-4)(-5)/3! x³

= 1 - 3x + 6x² - 10x³

With x = 0.02:

= 1 - 0.06 + 6(0.0004) - 10(0.000008)

= 1 - 0.06 + 0.0024 - 0.00008

= 0.94232

So 1/(2.04)³ ≈ (1/8) x 0.94232 = 0.11779

Calculator: 1/(2.04)³ = 0.117793...

Determining Accuracy

To assess accuracy, compare successive partial sums or examine the size of the next omitted term.

Worked Example 6

The expansion of (1 + x)^(1/2) is used with x = 0.1 to approximate sqrt(1.1).

Computing term by term:

n = 1/2, x = 0.1
Term 1: 1
Term 2: (1/2)(0.1) = 0.05
Term 3: (1/2)(-1/2)/2 x (0.1)² = -0.00125
Term 4: (1/2)(-1/2)(-3/2)/6 x (0.1)³ = +0.0000625

Terms used	Approximation	Calculator value	Error
Up to x	1.05	1.04881	1.19 x 10^(-3)
Up to x²	1.04875	1.04881	5.9 x 10^(-5)
Up to x³	1.0490125	1.04881	2.0 x 10^(-4)

The more terms we use and the smaller |x| is, the better the approximation. For very high accuracy, choose x to be as small as possible.

Choosing the Right Form

Worked Example 7

Find an approximation for sqrt(99) by choosing a suitable base.

Method 1: 99 = 100(1 - 0.01), so sqrt(99) = 10(1 - 0.01)^(1/2)

(1 - 0.01)^(1/2) ≈ 1 + (1/2)(-0.01) - (1/8)(0.0001)

= 1 - 0.005 - 0.0000125

= 0.9949875

sqrt(99) ≈ 10 x 0.9949875 = 9.949875

Method 2: 99 = 81(1 + 18/81) = 81(1 + 2/9)

sqrt(99) = 9(1 + 2/9)^(1/2), but x = 2/9 ≈ 0.222 which is much larger.

Method 1 gives better accuracy with fewer terms.

General principle: Choose the base to make x as small as possible.

Combining Expansions

Worked Example 8

Find the first three terms of the expansion of sqrt((1 + 3x)/(1 - x)).

sqrt((1 + 3x)/(1 - x)) = (1 + 3x)^(1/2) x (1 - x)^(-1/2)

Expand each:

(1 + 3x)^(1/2) ≈ 1 + (3/2)x + (1/2)(-1/2)/2 x (3x)² = 1 + (3/2)x - (9/8)x²

(1 - x)^(-1/2) ≈ 1 + (1/2)x + (-1/2)(-3/2)/2 x x² = 1 + (1/2)x + (3/8)x²

Multiply (keeping terms up to x²):

= (1 + (3/2)x - (9/8)x²)(1 + (1/2)x + (3/8)x²)

= 1 + (1/2)x + (3/8)x² + (3/2)x + (3/4)x² - (9/8)x²

= 1 + 2x + (3/8 + 6/8 - 9/8)x²

= 1 + 2x (the x² coefficient is zero)

Validity: |3x| < 1 and |x| < 1, so |x| < 1/3.

Percentage Error

Worked Example 9

The binomial expansion of (1 - x)^(1/3) up to x² is used with x = 0.1 to estimate (0.9)^(1/3). Find the percentage error.

(1 - x)^(1/3) ≈ 1 - x/3 - x²/9

With x = 0.1: 1 - 0.1/3 - 0.01/9 = 1 - 0.03333 - 0.00111 = 0.96556

Calculator: 0.9^(1/3) = 0.96549...

Percentage error = |0.96556 - 0.96549| / 0.96549 x 100% = 0.0073%

Exam Tips

Tip	Detail
Choose the base wisely	The closer x is to zero, the fewer terms needed for accuracy
State the expansion	Always write down the general expansion you are using before substituting
Show substitution	Write out each term separately before combining
Validity	Always state the range of x for which the expansion is valid
Perfect squares/cubes	Express the number as (perfect power)(1 + small fraction)
Accuracy	More terms and smaller x give better approximations

Practice Problems

Use binomial expansion to estimate sqrt(1.06) to 5 d.p. [Answer: 1.02956]
Estimate the cube root of 65 using a binomial expansion. [Answer: 4.0207]
Estimate 1/sqrt(3.96) using (4 - 0.04)^(-1/2). [Answer: 0.50125]
Find the percentage error when (1 + 0.05)^(-2) is approximated by 1 - 0.1 + 0.015. [Answer: about 0.012%]
Expand (4 - x)^(1/2) up to x² and use it to estimate sqrt(3.9). [Answer: 1.97484]

Summary

Binomial expansions approximate values by choosing a suitable base and small x.
Factor out a perfect power to write as (constant)(1 + small value)^n.
Smaller |x| means faster convergence and better accuracy.
Always state the expansion used, the substitution made, and the validity condition.
Compare with calculator values to assess accuracy.

A-Level Deep Dive: Binomial Estimation

Spec mapping

Edexcel 9MA0-02 specification, section 4 — Sequences and series, sub-strand on the binomial expansion for general $n$ covers the expansion $(1 + x)^n$ for rational $n$ , including its use for approximation; understanding that the expansion is valid for $|x| < 1$ (refer to the official specification document for exact wording). Binomial estimation is a Year 2 Pure topic that recombines content from Year 1 (the integer- $n$ binomial theorem) with new convergence reasoning. It appears most often on 9MA0-02 Paper 2, but the underlying ideas surface synoptically in section 7 (Differentiation, where the linear approximation $f(x) \approx f(a) + f'(a)(x - a)$ is exactly the first two terms of a binomial-style expansion), section 8 (Integration, where small- $x$ expansions generate convergent series for $\ln(1 + x)$ and similar), and section 5 (Trigonometry, via small-angle approximations $\sin x \approx x$ and $\cos x \approx 1 - x^2/2$ , which are the binomial-style expansions of trig functions to two terms). The Edexcel formula booklet does list the general binomial expansion — but only for $(1 + x)^n$ . Expansions of $(a + bx)^n$ require manual factoring, which is where most marks are won or lost.

Worked example with full mark scheme

Question (8 marks):

(a) Find the binomial expansion of $(1 + x)^{1/3}$ in ascending powers of $x$ , up to and including the term in $x^3$ , simplifying each coefficient. State the range of values of $x$ for which the expansion is valid. (5)

(b) Hence, by choosing a suitable value of $x$ , find an estimate for $\sqrt[3]{1.06}$ to 4 decimal places. (3)

Solution with mark scheme:

(a) Step 1 — apply the general binomial formula.

For $(1 + x)^n$ the expansion is

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

With $n = \tfrac{1}{3}$ :

$(1 + x)^{1/3} = 1 + \tfrac{1}{3}x + \dfrac{\tfrac{1}{3}\cdot(-\tfrac{2}{3})}{2}x^2 + \dfrac{\tfrac{1}{3}\cdot(-\tfrac{2}{3})\cdot(-\tfrac{5}{3})}{6}x^3 + \cdots$

M1 — applying the general binomial formula with the correct fractional $n$ . A common slip is to use $n = 3$ (the denominator of the index) or to forget that successive numerators decrease by 1.

Step 2 — simplify the coefficients.

Coefficient of $x^2$ : $\dfrac{(1/3)(-2/3)}{2} = \dfrac{-2/9}{2} = -\dfrac{1}{9}$ .
Coefficient of $x^3$ : $\dfrac{(1/3)(-2/3)(-5/3)}{6} = \dfrac{10/27}{6} = \dfrac{10}{162} = \dfrac{5}{81}$ .

So the expansion is

$(1 + x)^{1/3} = 1 + \tfrac{1}{3}x - \tfrac{1}{9}x^2 + \tfrac{5}{81}x^3 + \cdots$

A1 — coefficient of $x^2$ correct.

A1 — coefficient of $x^3$ correct.

Step 3 — state validity.

The general binomial expansion converges for $|x| < 1$ .

B1 — validity stated as $|x| < 1$ (or equivalently $-1 < x < 1$ ). Examiners reject "x < 1" alone and "x is small" — both lose this independent B mark.

(b) Step 1 — choose a suitable substitution.

To estimate $\sqrt[3]{1.06} = (1 + 0.06)^{1/3}$ , set $x = 0.06$ .

M1 — choosing $x = 0.06$ (or any equivalent value that makes the LHS equal $\sqrt[3]{1.06}$ ). The choice must lie within the validity range $|x| < 1$ , which $0.06$ does.

Step 2 — evaluate the expansion.

$(1 + 0.06)^{1/3} \approx 1 + \tfrac{1}{3}(0.06) - \tfrac{1}{9}(0.06)^2 + \tfrac{5}{81}(0.06)^3$

Term by term:

$\tfrac{1}{3}(0.06) = 0.02$
$\tfrac{1}{9}(0.0036) = 0.0004$
$\tfrac{5}{81}(0.000216) \approx 0.0000133$

Summing: $1 + 0.02 - 0.0004 + 0.0000133 \approx 1.0196133$ .

M1 — substituting and computing all four terms.

A1 — $\sqrt[3]{1.06} \approx 1.0196$ to 4 decimal places.

Total: 8 marks (M2 A3 B1 M1 A1).

A calculator gives $\sqrt[3]{1.06} = 1.019613\ldots$ , so the four-term truncation is accurate to all four decimal places — a useful sanity check that the validity condition is comfortably satisfied.

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): $f(x) = (4 - x)^{1/2}$ for $|x| < 4$ .

(a) Show that $f(x)$ can be written in the form $2(1 - \tfrac{x}{4})^{1/2}$ . (1)

(b) Hence find the binomial expansion of $f(x)$ in ascending powers of $x$ up to and including the term in $x^2$ , simplifying each coefficient. (3)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1a) — factoring out $4^{1/2} = 2$ to obtain $2(1 - x/4)^{1/2}$ .

(b)

M1 (AO1.1b) — applying the binomial expansion to $(1 - x/4)^{1/2}$ with $n = 1/2$ and the inner variable $-x/4$ .
A1 (AO1.1b) — coefficient of $x$ correct: $2 \cdot \tfrac{1}{2} \cdot (-\tfrac{1}{4}) = -\tfrac{1}{4}$ , giving the term $-\tfrac{x}{4}$ .
A1 (AO2.5) — coefficient of $x^2$ correct: $2 \cdot \dfrac{(1/2)(-1/2)}{2} \cdot \dfrac{1}{16} = -\dfrac{1}{64}$ , giving the term $-\tfrac{x^2}{64}$ .

(c)

M1 (AO3.1a) — choosing $x = 0.04$ so that $f(x) = (4 - 0.04)^{1/2} = \sqrt{3.96}$ , with $|x| = 0.04 < 4$ inside validity.
A1 (AO1.1b) — evaluating $2 - \tfrac{0.04}{4} - \tfrac{0.0016}{64} \approx 2 - 0.01 - 0.000025 = 1.989975$ .

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This question type rewards the candidate who treats factoring as a deliberate step rather than rushing into the expansion. The "show that" command in (a) is worth a free mark for any candidate who pauses to write the factoring out explicitly.

Synoptic links

Connects to:

Section 4 — General binomial expansion: the estimation use case is built directly on the expansion of $(1 + x)^n$ for rational $n$ . The convergence condition $|x| < 1$ is the defining feature that distinguishes the general binomial from the integer- $n$ case (which is finite and always valid). Estimation forces engagement with that condition rather than letting it stay abstract.
Taylor series (Year 2 / first-year university): the binomial expansion is a special case of Taylor's theorem applied to $f(x) = (1 + x)^n$ at the point $x = 0$ . The error after truncating at the $k$ th term is bounded by the next term in size when the series is alternating — which is exactly why four terms suffice to estimate $\sqrt[3]{1.06}$ to four decimal places.
Section 10 — Numerical methods: binomial estimation is the algebraic cousin of the Newton-Raphson iteration. Both produce successively better approximations, but the binomial expansion gives the entire approximation in closed form (a polynomial), whereas Newton-Raphson is recursive. Comparing the two on the same target — say $\sqrt[3]{1.06}$ — sharpens understanding of what "convergence" means.
Section 5 — Small-angle approximations: $\sin x \approx x$ and $\cos x \approx 1 - x^2/2$ are not binomial expansions, but they share the same logical structure: a function evaluated near a known value via the leading terms of its Taylor series. Recognising the parallel structure makes both topics easier to remember.
Floating-point arithmetic (computer science / numerical analysis): every time a calculator computes $\sqrt[3]{1.06}$ , an internal algorithm — often based on a truncated series expansion or Newton iteration — is running. The IEEE 754 double-precision standard guarantees about 16 significant figures of accuracy; the binomial expansion to four terms guarantees about four. Both are bounded approximations.

Mark-scheme literacy

Binomial estimation questions on 9MA0-02 split AO marks more evenly than pure procedural topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying the general binomial formula, simplifying coefficients, evaluating the expansion at a chosen $x$
AO2 (reasoning / interpretation)	25–35%	Selecting a suitable substitution, justifying that the chosen $x$ lies inside the validity range, factoring $(a + bx)^n$ correctly to expose $(1 + \text{small})^n$ form
AO3 (problem-solving)	10–20%	Choosing the optimal substitution to maximise accuracy with minimum computation, comparing truncation error to the requested precision

Examiner-rewarded phrasing: "the expansion is valid for $|x| < 1$ "; "since $|0.06| < 1$ , the substitution lies within the validity range"; "to estimate $\sqrt[3]{1.06}$ we choose $x = 0.06$ so that $(1 + x)^{1/3} = (1.06)^{1/3}$ ". Phrases that lose marks: "the expansion converges quickly" (vague — the B mark needs the symbolic condition); "x must be small" (insufficiently precise); using the substitution without writing it down ("by inspection" is not a substitution statement).

Binomial Estimation

Binomial Estimation

Recap: The General Binomial Expansion

Approximating Square Roots

Worked Example 1

Worked Example 2

Approximating Cube Roots

Worked Example 3

Approximating Reciprocals

Worked Example 4

Worked Example 5

Determining Accuracy

Worked Example 6

Choosing the Right Form

Worked Example 7

Combining Expansions

Worked Example 8

Percentage Error

Worked Example 9

Exam Tips

Practice Problems

Summary

A-Level Deep Dive: Binomial Estimation

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

More in Mathematics