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Pure content (sections A–J) · Subtopic 35 of 66

Binomial Expansion for Any Rational n

Extending (1+x)n(1+x)^n to negative and fractional powers: an infinite series that only works for ∣x∣<1|x|<1, how to handle (a+bx)n(a+bx)^n, and how to use the expansion with partial fractions and for approximations.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Expand (1+x)n(1+x)^n for negative and fractional nn up to a given power.
  • State the range of xx for which the expansion is valid.
  • Expand (a+bx)n(a+bx)^n by first taking out ana^n.
  • Combine expansions with products and partial fractions, and use them to approximate.
Pause and think

Long-divide 11 by (1+x)(1+x), or check: what is (1+x)(1−x+x2−x3+… )(1+x)(1-x+x^2-x^3+\dots)?

Section 2 of 14

Core idea

Start hereSquare roots by hand

How would you find 1.02\sqrt{1.02} without a calculator? It's (1+0.02)1/2(1+0.02)^{1/2}. Newton discovered that the binomial expansion works for fractional and negative powers too, as an infinite series, and the first few terms give superb approximations: 1.02≈1+0.01−0.00005=1.00995\sqrt{1.02}\approx1+0.01-0.00005=1.00995 (true value 1.009950…1.009950\ldots).

The big idea

For any rational nn and ∣x∣<1|x|<1: (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots. For (a+bx)n(a+bx)^n, first factor out ana^n.

1. The general expansion

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯ ,∣x∣<1(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots,\quad|x|<1
where
nn
any rational number (negative or fraction allowed).
∣x∣<1|x|<1
the validity condition: the series only converges for these x.

For a positive whole nn, a factor (n−n)=0(n-n)=0 eventually appears and the series stops. For other nn it never stops, and like a geometric series it only settles when the terms shrink, i.e. ∣x∣<1|x|<1.

(1 + x)⁻¹ and its approximations
−1.5−1−0.50.511.5−112340xyy = 1/(1 + x)
Between the dashed lines (−1<x<1-1<x<1) adding more terms brings the polynomial closer to the curve. Outside that range, extra terms make it worse: the series only works for ∣x∣<1|x|<1.
Worked example(1+x)−2(1+x)^{-2} up to x3x^3

Expand.

Goal
Four terms and validity.
Strategy
n=−2n=-2.
  1. 1+(−2)x+(−2)(−3)2x2+(−2)(−3)(−4)6x31+(-2)x+\frac{(-2)(-3)}{2}x^2+\frac{(-2)(-3)(-4)}{6}x^3
    Substitute n=−2n=-2.
  2. =1−2x+3x2−4x3=1-2x+3x^2-4x^3
    Simplify.
Answer1−2x+3x2−4x3+⋯1-2x+3x^2-4x^3+\cdots, ∣x∣<1|x|<1.
✓ Checkx=0.1x=0.1: 1−0.2+0.03−0.004=0.8261-0.2+0.03-0.004=0.826; 1.1−2=0.82641.1^{-2}=0.8264 ✓.
Worked example1−4x\sqrt{1-4x} up to x2x^2

Expand and state validity.

Goal
Series.
Strategy
n=12n=\frac12, replace xx by −4x-4x (bracket it!).
  1. 1+12(−4x)+12(−12)2(−4x)21+\tfrac12(-4x)+\frac{\frac12(-\frac12)}{2}(-4x)^2
    Substitute.
  2. =1−2x−18(16x2)=1−2x−2x2=1-2x-\tfrac18(16x^2)=1-2x-2x^2
    (−4x)2=16x2(-4x)^2=16x^2.
  3. ∣−4x∣<1⇒∣x∣<14|-4x|<1\Rightarrow|x|<\tfrac14
    Validity.
Answer1−2x−2x21-2x-2x^2, ∣x∣<14|x|<\frac14.
✓ Checkx=0.01x=0.01: 0.97980.9798; 0.96=0.97980\sqrt{0.96}=0.97980 ✓.

2. When the first term isn't 1

Worked example(8+3x)1/3(8+3x)^{1/3} up to x2x^2

Expand.

Goal
Series and validity.
Strategy
Factor 88 out so the bracket starts with 11.
  1. (8+3x)1/3=81/3(1+38x)1/3=2(1+38x)1/3(8+3x)^{1/3}=8^{1/3}\left(1+\tfrac38x\right)^{1/3}=2\left(1+\tfrac38x\right)^{1/3}
    8+3x=8(1+38x)8+3x=8(1+\frac38x); index laws.
  2. =2[1+13⋅38x+13(−23)2(38x)2]=2\left[1+\tfrac13\cdot\tfrac38x+\frac{\frac13(-\frac23)}{2}\left(\tfrac38x\right)^2\right]
    Expand with n=13n=\frac13.
  3. =2[1+18x−164x2]=2+14x−132x2=2\left[1+\tfrac18x-\tfrac1{64}x^2\right]=2+\tfrac14x-\tfrac1{32}x^2
    −19×964=−164-\frac19\times\frac9{64}=-\frac1{64}.
  4. ∣38x∣<1⇒∣x∣<83\left|\tfrac38x\right|<1\Rightarrow|x|<\tfrac83
    Validity.
Answer2+14x−132x22+\frac14x-\frac1{32}x^2, ∣x∣<83|x|<\frac83.
✓ Checkx=0x=0: 83=2\sqrt[3]8=2 ✓.
Common misconceptionForgetting the power on the factor (8+3x)1/3=81/3(…)(8+3x)^{1/3}=8^{1/3}(\ldots), i.e. 2(…)2(\ldots), not 8(…)8(\ldots).
Edge caseCombining with partial fractions Expand each partial fraction; the overall validity is the stricter condition.
Pause and think

Estimate 1.02\sqrt{1.02} with three terms.

Summary

  • General binomial series, valid ∣x∣<1|x|<1.
  • Bracket substituted terms.
  • Factor out ana^n first.
Key words
Validity
Range of x where the series converges.
HistoryNewton's discovery. Around 1665, while Cambridge was closed because of the plague, the young Isaac Newton discovered that the binomial expansion works for fractional and negative powers. He used it to calculate square roots and areas to many decimal places, and it became a key tool in his calculus.
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