Expand rational powers of one plus x with correct coefficients
Extend the binomial expansion to rational exponents, keeping signs, factorial divisors and its strict validity condition.
Section 3.1 · Algebra
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.1. Paper 1 knowledge is assumed. Shared modulus, polynomial division, factor and remainder lessons are linked from both sections 2.1 and 3.1. Paper 3 adds partial fractions and rational-power binomial expansions. Partial-fraction numerators may have the same degree as the denominator, but higher-degree numerators are excluded. General-term questions for the rational binomial expansion are not included. Use the stated strict validity condition. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Expand a positive-integer binomial and calculate factorials.
- Work with signed fractions, index notation and powers of a small input.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 3!?
What you will learn
- Use the first terms of (1 + x)ⁿ for rational n, including negative integers and fractional powers.
- Distinguish a truncated approximation from an exact polynomial and state |x| < 1.
Use descending exponent factors and factorial divisors
For rational n and |x| < 1, the expansion begins (1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + … . The successive exponent factors continue to decrease by 1; they are not all equal to n. The factorial divisors are 2, 6, 24 and so on. For a non-negative integer n the expansion eventually terminates, but a negative integer or non-integer rational exponent generally gives an infinite series.
For n = −2, the coefficients through x³ are 1, −2, 3 and −4: (1 + x)⁻² = 1 − 2x + 3x² − 4x³ + … , valid for |x| < 1. The x² coefficient comes from (−2)(−3)/2 = 3; the x³ coefficient comes from (−2)(−3)(−4)/6 = −4. Retain the alternating signs rather than copying a positive-integer Pascal row.
Keep the series condition and approximation separate
For n = 1/2, the first four terms are 1 + x/2 − x²/8 + x³/16. Thus √(1 + x) is approximately this cubic for small x within |x| < 1, but the cubic is not the exact square-root function. At x = −3/4, the exact square root is 1/2 and the cubic is 541/1024, approximately 0.52832. The visible difference is a reminder that being inside the validity interval alone does not guarantee a requested numerical accuracy after only four terms.
Write an ellipsis or an explicit truncation when reporting an infinite expansion. If asked for terms up to x³, include every term through that power and state |x| < 1. The strict condition describes the syllabus's guaranteed expansion interval; it does not say that the original function is undefined at every endpoint or beyond the interval. Finding a general term for these rational expansions is outside the stated requirements.
Pause and explain
Which coefficient multiplies x² in (1 + x)¹ᐟ²?
Worked example
Expand (1 + x)⁻² through the x³ term and state the condition.
Show the worked solution
- Substitute n = −2 into the descending-factor formula.
- Compute coefficients −2, (−2)(−3)/2 = 3, and (−2)(−3)(−4)/6 = −4.
- Retain the ellipsis because this negative-integer expansion is infinite, with |x| < 1.
Answer 1 − 2x + 3x² − 4x³ + … , valid for |x| < 1.
Try a topic test
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Expand (1 + x)⁻¹ through x³ and state the validity interval.
Give me a hint
Use exponent −1 with factors −1, −2 and −3, keeping the factorial denominators.
Compare my reasoning
- The linear coefficient is −1 and the quadratic coefficient is (−1)(−2)/2 = 1.
- The cubic coefficient is (−1)(−2)(−3)/6 = −1.
- The expansion is 1 − x + x² − x³ + … , for −1 < x < 1; the negative exponent does not make the series terminate.
1 − x + x² − x³ + … , |x| < 1.
Look for these in your work
- Compute signs and factorial divisors correctly.
- State the strict interval and indicate omitted higher powers.
Expand (1 + x)⁻¹ᐟ² through x³.
Give me a hint
Use n = −1/2 and subtract 1 from each successive exponent factor.
Compare my reasoning
- The linear coefficient is −1/2; the quadratic coefficient is (−1/2)(−3/2)/2 = 3/8.
- The cubic coefficient is (−1/2)(−3/2)(−5/2)/6 = −5/16.
- Write 1 − x/2 + 3x²/8 − 5x³/16 + … , keeping the condition |x| < 1.
1 − x/2 + 3x²/8 − 5x³/16 + … , |x| < 1.
Look for these in your work
- Use the descending fractional factors.
- Keep exact coefficients and the validity condition.
Expand (1 + x)²ᐟ³ through x³, explaining why the quadratic and cubic coefficients have different signs.
Give me a hint
The descending factors after 2/3 are −1/3 and −4/3.
Compare my reasoning
- The linear coefficient is 2/3, and (2/3)(−1/3)/2 gives −1/9 for x².
- The two negative factors make (2/3)(−1/3)(−4/3)/6 = 4/81 positive for x³.
- Thus the expansion is 1 + 2x/3 − x²/9 + 4x³/81 + … , valid for |x| < 1.
1 + 2x/3 − x²/9 + 4x³/81 + … , |x| < 1.
Look for these in your work
- Derive each sign from the actual exponent factors.
- Report all four terms as an infinite-series truncation.
Common mistakes
- Decrease the exponent factor at each stage; n³ is not the cubic numerator.
- Include the factorial divisor and all signs.
- A finite truncation is not exactly equal to the full function for arbitrary non-zero x.
What can you explain now?
Find the coefficient of x² in √(1 + x).
Compare with the explanation
−1/8.
With n = 1/2, n(n − 1)/2 = (1/2)(−1/2)/2 = −1/8.
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Make it stick
Derive the first four coefficients for a rational exponent, explain which integer powers terminate, and separate validity from the accuracy of a short truncation.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.