Combine binomial expansions and check a numerical approximation
Collect all contributions to each requested power, intersect validity intervals and verify a truncated estimate against the original expression.
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
- Adapt a rational-power binomial expansion with its validity interval.
- Multiply polynomials, collect like powers and decompose simple rational fractions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which products contribute to x² when (1 + 2x) multiplies 1 + x/2 + 3x²/8 + … ?
What you will learn
- Multiply adapted binomial series or use partial fractions to expand a rational expression through a requested power.
- Intersect the component validity intervals and distinguish an estimate from a guaranteed accurate or exact result.
Each coefficient may receive more than one contribution
To expand (1 + 2x)/√(1 − x), first use (1 − x)⁻¹ᐟ² = 1 + x/2 + 3x²/8 + 5x³/16 + … . Multiplication by 1 + 2x gives 1 + 5x/2 + 11x²/8 + 17x³/16 + … , valid for |x| < 1. The quadratic term receives 3x²/8 from the constant multiplier and x² from 2x times x/2. The cubic receives 5x³/16 and 3x³/4. Do not stop after taking the highest-power term from just one factor.
For √[(1 + x)/(1 − x)] near zero, multiply (1 + x)¹ᐟ² and (1 − x)⁻¹ᐟ². Both expansions require |x| < 1, where both real factors are valid. Collecting terms gives 1 + x + x²/2 + x³/2 + … . The original ratio may have a positive square root in other real regions, but these local factors and series are being used only on their stated common interval.
Partial fractions can simplify an expansion; use the common interval
For 1/[(1 − x)(1 + 2x)], the identity is 1/[3(1 − x)] + 2/[3(1 + 2x)]. The first expansion requires |x| < 1; the second requires |2x| < 1. Their common condition is |x| < 1/2. Combining the series gives 1 − x + 3x² − 5x³ + … . At x = 3/5 the original fraction exists, but the second component is outside its binomial interval, so this expansion cannot be justified there.
At x = 1/10 the cubic estimate is 1 − 0.1 + 0.03 − 0.005 = 0.925. The original value is 1/[(9/10)(6/5)] = 25/27, approximately 0.925926. The estimate underestimates by 1/1080, approximately 0.000926. This difference comes from checking the original expression; a small input or a valid expansion interval alone is not a bound on the omitted terms. Use the requested number of terms and report an approximation with appropriate notation.
Pause and explain
For 1/[(1 − x)(1 + 2x)], which condition justifies combining both binomial expansions?
Worked example
Expand (1 + 2x)/√(1 − x) through x³.
Show the worked solution
- Use (1 − x)⁻¹ᐟ² = 1 + x/2 + 3x²/8 + 5x³/16 + … .
- Multiply by 1 + 2x and combine contributions: 1/2 + 2 = 5/2, 3/8 + 1 = 11/8, and 5/16 + 3/4 = 17/16.
- Retain terms through x³, the ellipsis and the common interval |x| < 1.
Answer 1 + 5x/2 + 11x²/8 + 17x³/16 + … , |x| < 1.
Try a topic test
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Test this topic →From guided practice to a new situation
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 + x)⁻² through x³.
Give me a hint
Multiply 1 − x by the first four terms of the exponent −2 expansion.
Compare my reasoning
- Use (1 + x)⁻² = 1 − 2x + 3x² − 4x³ + … .
- Collect coefficients: −2 − 1 = −3, 3 + 2 = 5, and −4 − 3 = −7.
- The result is 1 − 3x + 5x² − 7x³ + … , with |x| < 1.
1 − 3x + 5x² − 7x³ + … , |x| < 1.
Look for these in your work
- Include contributions from both multiplier terms.
- Retain the original strict binomial interval.
Expand √[(1 + x)/(1 − x)] through x³ near x = 0.
Give me a hint
Multiply the square-root series for 1 + x by the reciprocal-square-root series for 1 − x.
Compare my reasoning
- Use 1 + x/2 − x²/8 + x³/16 and 1 + x/2 + 3x²/8 + 5x³/16.
- The linear coefficient is 1; the quadratic is −1/8 + 1/4 + 3/8 = 1/2; the cubic is 1/16 − 1/16 + 3/16 + 5/16 = 1/2.
- Both factors and expansions are valid on |x| < 1, giving 1 + x + x²/2 + x³/2 + … .
1 + x + x²/2 + x³/2 + … , |x| < 1.
Look for these in your work
- Collect all cross-products through the cubic power.
- State the common local interval for both real factors.
Use partial fractions to expand 1/[(1 − x)(1 + 2x)] through x³. Estimate its value at x = 0.1, compare with the original value, and explain why x = 0.6 is outside this expansion's interval.
Give me a hint
Decompose first, intersect the two strict conditions and compare the cubic value with the exact fraction.
Compare my reasoning
- The partial fractions are 1/[3(1 − x)] + 2/[3(1 + 2x)], giving 1 − x + 3x² − 5x³ + … on |x| < 1/2.
- At 0.1 the cubic is 0.925, while the original value is 25/27 ≈ 0.925926; the underestimate is 1/1080 ≈ 0.000926.
- At 0.6 the original fraction is defined, but |2x| = 1.2 > 1, so its second binomial expansion is outside the required interval.
1 − x + 3x² − 5x³ + … , |x| < 1/2; estimate 0.925 at 0.1, underestimate 1/1080; 0.6 is outside the series interval.
Look for these in your work
- Combine exact partial fractions and intersect their expansion conditions.
- Check the numerical estimate against the original expression without claiming unverified accuracy.
Common mistakes
- Collect every product contributing to a requested power.
- Use the intersection of validity intervals, not their union.
- A valid truncated series is an estimate; do not claim exactness or unverified decimal accuracy.
What can you explain now?
What is the common interval for |x| < 1 and |2x| < 1?
Compare with the explanation
|x| < 1/2.
Both conditions must hold simultaneously; the second imposes the narrower bound.
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Make it stick
Expand a product or rational expression through a specified power, justify its common interval and explain how checking the original value assesses a truncated estimate.
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.