Binomial Expansion for General (Rational) n

When n is not a positive integer — for example n = 1/2, n = -1, or n = -3/2 — we can still expand (1 + x)^n, but the expansion is an infinite series and is only valid for |x| < 1. This extends the binomial theorem and is tested at A-Level (9MA0).

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The General Binomial Expansion

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

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

This is an infinite series (it never terminates unless n is a non-negative integer).

Key condition: the expansion is valid only when |x| < 1.

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How This Differs from Positive Integer n

Feature	Positive integer n	General/rational n
Number of terms	Finite (n + 1 terms)	Infinite
Values of x	All real x	Only
Coefficient formula	n!/(r!(n-r)!)	n(n-1)(n-2)...(n-r+1)/r!
Result	Exact	Approximation (using first few terms)

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Expanding (1 + x)^n for Rational n

Worked Example 1

Find the expansion of (1 + x)^(1/2) up to and including the term in x³.

Using the formula with n = 1/2:

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

= 1 + (1/2)x + (-1/4)/2 x² + (3/8)/6 x³ + ...

= 1 + (1/2)x - (1/8)x² + (1/16)x³ + ...

Valid for |x| < 1

Worked Example 2

Find the expansion of (1 + x)^(-1) up to the term in x³.

With n = -1:

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

= 1 - x + x² - x³ + ...

Valid for |x| < 1

This is the well-known geometric series formula: 1/(1 + x) = 1 - x + x² - x³ + ...

Worked Example 3

Find the expansion of (1 + x)^(-2) up to and including the x³ term.

With n = -2:

= 1 + (-2)x + (-2)(-3)/2! x² + (-2)(-3)(-4)/3! x³ + ...

= 1 - 2x + 3x² - 4x³ + ...

Valid for |x| < 1

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Expanding (a + bx)^n

When the expression is not in the form (1 + x)^n, factor out to create it.

(a + bx)^n = a^n x (1 + bx/a)^n

Then expand (1 + bx/a)^n and multiply by a^n.

The validity condition becomes |bx/a| < 1, i.e. |x| < |a/b|.

Worked Example 4

Expand (4 + x)^(1/2) up to the x² term and state the range of validity.

= (4)^(1/2) x (1 + x/4)^(1/2)

= 2 x [1 + (1/2)(x/4) + (1/2)(-1/2)/2! (x/4)² + ...]

= 2 x [1 + x/8 - x²/128 + ...]

= 2 + x/4 - x²/64 + ...

Validity: |x/4| < 1, so |x| < 4

Worked Example 5

Expand 1/(2 - 3x)² up to the x² term and state the validity.

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

= (1/4) x (1 + (-3x/2))^(-2)

Let u = -3x/2. Using the expansion of (1 + u)^(-2) = 1 - 2u + 3u² - ...

= (1/4) x [1 - 2(-3x/2) + 3(-3x/2)² - ...]

= (1/4) x [1 + 3x + 27x²/4 - ...]

= 1/4 + 3x/4 + 27x²/16 + ...

Validity: |3x/2| < 1, so |x| < 2/3

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Using Expansions to Approximate Values

Worked Example 6

Use the expansion of (1 + x)^(1/2) to estimate sqrt(1.04).

From Example 1: (1 + x)^(1/2) ≈ 1 + (1/2)x - (1/8)x²

With x = 0.04:

≈ 1 + (1/2)(0.04) - (1/8)(0.04)²

= 1 + 0.02 - (1/8)(0.0016)

= 1 + 0.02 - 0.0002

= 1.0198

(Calculator gives sqrt(1.04) = 1.01980390..., so this is accurate to 4 d.p.)

Worked Example 7

Estimate the cube root of 8.24 using a binomial expansion.

8.24 = 8(1 + 0.03)

Cube root of 8.24 = (8)^(1/3) x (1 + 0.03)^(1/3)

= 2 x [1 + (1/3)(0.03) + (1/3)(-2/3)/2! (0.03)² + ...]

= 2 x [1 + 0.01 - (1/9)(0.0009) + ...]

= 2 x [1 + 0.01 - 0.0001 + ...]

= 2 x 1.0099

= 2.0198

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Partial Fractions and Binomial Expansion

A common A-Level question combines partial fractions with binomial expansion.

Worked Example 8

Express (3x + 1) / ((1 + x)(1 - 2x)) in partial fractions and hence find the expansion up to x².

Partial fractions: (3x + 1) / ((1 + x)(1 - 2x)) = A/(1 + x) + B/(1 - 2x)

3x + 1 = A(1 - 2x) + B(1 + x)

Let x = -1: -3 + 1 = A(3), so A = -2/3 Let x = 1/2: 3/2 + 1 = B(3/2), so B = 5/3

So: -2/3 x (1 + x)^(-1) + 5/3 x (1 - 2x)^(-1)

Expand each: (1 + x)^(-1) = 1 - x + x² - ... (1 - 2x)^(-1) = 1 + 2x + 4x² + ...

= -2/3 x (1 - x + x²) + 5/3 x (1 + 2x + 4x²)

= (-2/3 + 2x/3 - 2x²/3) + (5/3 + 10x/3 + 20x²/3)

= 1 + 4x + 6x²

Validity: |x| < 1 and |2x| < 1, so |x| < 1/2 (the stricter condition applies).

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Exam Tips

Tip	Detail
Always state validity	You must state the range of x for which the expansion is valid to gain full marks
Factor out first	For (a + bx)^n, always write as a^n(1 + bx/a)^n before expanding
Careful with signs	n(n-1) when n = -2 gives (-2)(-3) = 6, not -6
Check with substitution	Substitute a small x value to verify your expansion
Partial fractions	If the question involves a rational function, expect partial fractions first

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Practice Problems

1. Expand (1 + x)^(-3) up to x³. [Answer: 1 - 3x + 6x² - 10x³]
2. Expand (1 - 2x)^(1/2) up to x². [Answer: 1 - x - x²/2; valid for |x| < 1/2]
3. Expand (9 + x)^(1/2) up to x². [Answer: 3 + x/6 - x²/216; valid for |x| < 9]
4. Use a binomial expansion to estimate sqrt(0.98) to 4 d.p. [Answer: 0.9899]
5. Express 5/((1+x)(1-x)) in partial fractions and expand up to x². [Answer: 5 + 5x²]

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Summary

• For rational n, (1 + x)^n = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + ...
• Valid only for |x| < 1.
• For (a + bx)^n, factor out a^n and use the substitution x to bx/a.
• Validity becomes |x| < |a/b|.
• These expansions are used for approximating values and simplifying rational functions.

A-Level Deep Dive: Binomial Expansion (General/Rational n)

Spec mapping

Edexcel 9MA0-02 specification, Year 2 Pure Mathematics, sub-strand 4.4 covers the expansion of $(1 + x)^n$(1+x)n for any rational $n$n, including its use for approximation; be aware that the expansion is valid for $|x| < 1$∣x∣<1 (proof not required) (refer to the official specification document for exact wording). This sits inside Sequences and series and is examined in Paper 2 — Pure Mathematics, where it routinely carries 6–10 marks across one or two questions. The Year 2 expansion extends the Year 1 result: where the positive-integer binomial $(a + b)^n = \sum \binom{n}{k} a^{n-k} b^k$(a+b)n=∑(kn​)an−kbk terminates after $n+1$n+1 terms, the rational-$n$n version becomes an infinite series and only converges when $|x| < 1$∣x∣<1. The validity statement is itself an A-mark — omitting it costs marks even when every coefficient is correct.

The expansion in the formula booklet is given as:

$(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots, \quad |x| < 1$(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3+…,∣x∣<1

Synoptic reach is wide: positive-integer binomial (Year 1, sub-strand 4.3) provides the algebraic template; partial fractions (sub-strand 2.10) feed naturally into rational-$n$n expansion when each linear factor is rewritten as a power series; the Taylor / Maclaurin series of further-maths is the natural generalisation (the binomial is the Maclaurin series of $(1+x)^n$(1+x)n); integration of $(1+x)^{-1}$(1+x)−1 gives $\ln(1+x)$ln(1+x), providing a route to series for logarithms; and calculator-free approximation of expressions like $\sqrt{1.04}$1.04​ is the classic application question.

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Worked example with full mark scheme

Question (8 marks):

(a) Find the binomial expansion of $(1 - 2x)^{1/2}$(1−2x)1/2 in ascending powers of $x$x up to and including the term in $x^3$x3, simplifying each coefficient. State the range of values of $x$x for which the expansion is valid. (5)

(b) Hence, by substituting $x = 0.01$x=0.01, find an approximation to $\sqrt{0.98}$0.98​, giving your answer to 6 decimal places. (3)

Solution with mark scheme:

(a) Step 1 — identify $n$n and the variable.

Here $n = \tfrac{1}{2}$n=21​ and the bracket variable is $(-2x)$(−2x), not $x$x. Substitute carefully into the formula $(1 + X)^n$(1+X)n with $X = -2x$X=−2x.

M1 — correct identification of $n = \tfrac{1}{2}$n=21​ and $X = -2x$X=−2x.

Step 2 — compute coefficients.

Constant term: $1$1.

Term in $X$X: $nX = \tfrac{1}{2}(-2x) = -x$nX=21​(−2x)=−x.

Term in $X^2$X2: $\dfrac{n(n-1)}{2!}X^2 = \dfrac{(1/2)(-1/2)}{2}(-2x)^2 = -\dfrac{1}{8} \cdot 4x^2 = -\dfrac{x^2}{2}$2!n(n−1)​X2=2(1/2)(−1/2)​(−2x)2=−81​⋅4x2=−2x2​.

Term in $X^3$X3: $\dfrac{n(n-1)(n-2)}{3!}X^3 = \dfrac{(1/2)(-1/2)(-3/2)}{6}(-2x)^3 = \dfrac{3/8}{6}(-8x^3) = \dfrac{1}{16}(-8x^3) = -\dfrac{x^3}{2}$3!n(n−1)(n−2)​X3=6(1/2)(−1/2)(−3/2)​(−2x)3=63/8​(−8x3)=161​(−8x3)=−2x3​.

M1 — correct coefficient structure $\dfrac{n(n-1)}{2!}$2!n(n−1)​ etc. with the signed factors of $n - k$n−k.

A1 — correct $x$x and $x^2$x2 coefficients: $-x$−x and $-\tfrac{1}{2}x^2$−21​x2.

A1 — correct $x^3$x3 coefficient: $-\tfrac{1}{2}x^3$−21​x3.

So $(1 - 2x)^{1/2} \approx 1 - x - \tfrac{1}{2}x^2 - \tfrac{1}{2}x^3$(1−2x)1/2≈1−x−21​x2−21​x3.

Step 3 — state the validity condition.

The expansion of $(1 + X)^n$(1+X)n for non-integer $n$n is valid for $|X| < 1$∣X∣<1. Here $X = -2x$X=−2x, so $|-2x| < 1 \implies |x| < \tfrac{1}{2}$∣−2x∣<1⟹∣x∣<21​.

B1 — validity $|x| < \tfrac{1}{2}$∣x∣<21​ stated explicitly.

(b) Step 1 — engineer $1 - 2x = 0.98$1−2x=0.98.

Setting $1 - 2x = 0.98$1−2x=0.98 gives $x = 0.01$x=0.01, which lies inside $|x| < \tfrac{1}{2}$∣x∣<21​, so the expansion is valid here.

M1 — correct value of $x$x identified, with a comment confirming the substitution lies inside the validity range.

Step 2 — substitute.

$\sqrt{0.98} \approx 1 - 0.01 - \tfrac{1}{2}(0.01)^2 - \tfrac{1}{2}(0.01)^3$0.98​≈1−0.01−21​(0.01)2−21​(0.01)3 $= 1 - 0.01 - 0.00005 - 0.0000005$=1−0.01−0.00005−0.0000005 $= 0.9899495$=0.9899495

M1 — substitution carried out arithmetically with each term retained.

A1 — final answer $\sqrt{0.98} \approx 0.989949$0.98​≈0.989949 to 6 d.p.

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

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks):

$f(x) = (4 + x)^{1/2}$f(x)=(4+x)1/2.

(a) Find the binomial expansion of $f(x)$f(x) in ascending powers of $x$x up to and including the term in $x^2$x2, giving each coefficient as a simplified fraction. (4)

(b) State the range of values of $x$x for which the expansion is valid. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — factor out $4$4 correctly: $(4 + x)^{1/2} = 4^{1/2}(1 + x/4)^{1/2} = 2(1 + x/4)^{1/2}$(4+x)1/2=41/2(1+x/4)1/2=2(1+x/4)1/2.
• M1 (AO1.1b) — apply $(1 + X)^{1/2}$(1+X)1/2 with $X = x/4$X=x/4, $n = \tfrac{1}{2}$n=21​: constant $1$1, then $\tfrac{1}{2}(x/4)$21​(x/4), then $\dfrac{(1/2)(-1/2)}{2}(x/4)^2 = -\dfrac{1}{8} \cdot \dfrac{x^2}{16} = -\dfrac{x^2}{128}$2(1/2)(−1/2)​(x/4)2=−81​⋅16x2​=−128x2​.
• A1 (AO1.1b) — correct $x$x and $x^2$x2 coefficients in the bracket: $\tfrac{x}{8}$8x​ and $-\tfrac{x^2}{128}$−128x2​.
• A1 (AO2.5) — multiply the leading factor of $2$2 through cleanly: $f(x) \approx 2 + \tfrac{x}{4} - \tfrac{x^2}{64}$f(x)≈2+4x​−64x2​.

(b)

• M1 (AO1.1b) — set $|x/4| < 1$∣x/4∣<1.
• A1 (AO2.5) — solve to obtain $|x| < 4$∣x∣<4.

Total: 6 marks split AO1 = 4, AO2 = 2. The factoring step in (a) is the gateway: candidates who attempt to expand $(4 + x)^{1/2}$(4+x)1/2 directly without factoring out $4$4 first will write something like $\dfrac{1}{2} \cdot 4^{-1/2} \cdot x = \dfrac{x}{4}$21​⋅4−1/2⋅x=4x​ and may struggle to keep the structure correct on later terms. The "factor first" technique is the examiner-rewarded approach.

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Synoptic links

Connects to:

1. Sub-strand 4.3 — Positive-integer binomial: the Year 1 result $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$(a+b)n=∑k=0n​(kn​)an−kbk is the finite case. When $n$n is a positive integer, the factor $n - k$n−k in the numerator hits zero at $k = n$k=n and all subsequent terms vanish. For rational $n$n, no such termination occurs and the series is genuinely infinite, requiring the convergence condition $|x| < 1$∣x∣<1.

2. Sub-strand 2.10 — Partial fractions: an expression like $\dfrac{1}{(1-x)(1+2x)}$(1−x)(1+2x)1​ is split into partial fractions $\dfrac{A}{1-x} + \dfrac{B}{1+2x}$1−xA​+1+2xB​, each of which can be expanded as $(1-x)^{-1} = 1 + x + x^2 + \ldots$(1−x)−1=1+x+x2+… and $(1+2x)^{-1} = 1 - 2x + 4x^2 - \ldots$(1+2x)−1=1−2x+4x2−… using the rational binomial. The combined series gives a power-series expression for the original rational function.

3. Further Mathematics — Taylor / Maclaurin series: the Maclaurin expansion of $(1+x)^n$(1+x)n is $\sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!}x^k$∑k=0∞​k!f(k)(0)​xk, and computing $f^{(k)}(0)$f(k)(0) explicitly yields $\dfrac{n(n-1)\cdots(n-k+1)}{k!}x^k$k!n(n−1)⋯(n−k+1)​xk — exactly the binomial coefficient. The general binomial theorem is therefore the first non-trivial Maclaurin series students meet.

4. Sub-strand 8.2 — Integration of $(1+x)^{-1}$(1+x)−1: integrating the geometric series $1 - x + x^2 - x^3 + \ldots$1−x+x2−x3+… term-by-term gives $x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \ldots = \ln(1+x)$x−2x2​+3x3​−…=ln(1+x) for $|x| < 1$∣x∣<1. This route is how series for $\ln$ln and $\arctan$arctan are derived in undergraduate analysis.

5. Calculator-free approximation: substituting small $x$x values into a binomial expansion approximates roots and reciprocals ($\sqrt{1.02}$1.02​, $1/\sqrt{1.04}$1/1.04​, $\sqrt[3]{8.06}$38.06​). The accuracy depends on the truncation error, which the next term gives an upper bound on for alternating series.

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Mark-scheme literacy

Binomial-expansion (rational $n$n) questions on 9MA0-02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correctly applying the formula, computing coefficients, factoring out leading constants, simplifying surds and fractions
AO2 (reasoning / interpretation)	20–30%	Stating and justifying the validity range $\|x\| < r$∥x∥<r for the appropriate $r$r, presenting answers in the form requested, recognising why a substitution is or is not in range
AO3 (problem-solving)	5–15%	Combining partial fractions with binomial expansion, choosing a suitable $x$x for an approximation question, error estimation

Examiner-rewarded phrasing: "valid for $|x| < \tfrac{1}{2}$∣x∣<21​" (or whatever the bound is — always an explicit modulus inequality, not "for small $x$x"); "factor out $a$a first to obtain $a^n(1 + x/a)^n$an(1+x/a)n"; "since $|0.01| < \tfrac{1}{2}$∣0.01∣<21​, the substitution is within the range of validity". Phrases that lose marks: writing the validity as $x < \tfrac{1}{2}$x<21​ (missing the modulus, so negative $x$x values are wrongly excluded); stating "the series converges" without the explicit inequality (B0 — the bound is the mark); applying the positive-integer formula $\binom{n}{k}$(kn​) to a rational $n$n, which makes no sense as $\binom{1/2}{2}$(21/2​) is not a "choose" coefficient.

A specific Edexcel pattern: the validity condition is examined with two common pitfalls. First, when the bracket variable is a multiple of $x$x ($X = -2x$X=−2x, say), candidates often quote $|x| < 1$∣x∣<1 rather than transferring the bound through to obtain $|x| < \tfrac{1}{2}$∣x∣<21​. Second, when the bracket has been factored as $a^n(1 + x/a)^n$an(1+x/a)n, the bound becomes $|x/a| < 1 \implies |x| < |a|$∣x/a∣<1⟹∣x∣<∣a∣, which candidates routinely forget to scale up.

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Grade-band model answers

3-mark question

Question: Find the first three terms in the binomial expansion of $(1 + 3x)^{-2}$(1+3x)−2 in ascending powers of $x$x, simplifying each coefficient.

Grade C response (~210 words):