OCR A Level Mathematics A: Binomial expansion: integer and rational powers (H240) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, sections 1.04a to 1.04d and the partial fractions use in 1.02y. Questions 1 to 5 use positive integer powers and the probability link (Stage 1, common with AS). Questions 6 to 11 use rational powers or validity and are A Level only. Work every question by hand and show each step.

Links: OCR Mathematics A hub, printable checklist, study guide, revision notes and free diagnostics.

Questions

1. Given that nC2 = 78, find the value of n. [3]

2. Expand (3 − 4x)⁴ in ascending powers of x, simplifying each coefficient. [4]

3. (a) Find the coefficients of x² and x³ in the expansion of (3 − x)⁶. [2] (b) Hence find the coefficient of x³ in the expansion of (1 + 2x)(3 − x)⁶. [3]

4. The coefficient of x² in the expansion of (2 + kx)⁶ is 960. Find the possible values of the constant k. [4]

5. A player scores with each shot with probability 0.8, independently. She takes 6 shots. Let X be the number of scores. (a) Find P(X = 4), giving your answer to 3 significant figures. [2] (b) Find P(X ≥ 5), giving your answer to 3 significant figures. [3] (c) Explain why the six probabilities in the expansion of (0.2 + 0.8)⁶ must add to 1, and use P(X = 0) to find P(X ≥ 1). [2]

6. Expand (8 − 3x)^(−1/3) as a series up to and including the term in x³, and say for which x the series is valid. [6]

7. (a) Give the interval of x on which (3 − 2x)⁻² can be written as a power series. [2] (b) A student expands (5 + 2x)^(1/2) and writes that it is valid for |x| < 2/5. Identify the error and give the correct range. [2] (c) Explain why the expansion of (2 + x)⁶ has no restriction on x. [1]

8. (a) Expand (1 − 2x)^(1/2) up to and including the term in x³. [3] (b) Use your expansion with x = 0.03 to estimate √0.94, giving your answer to 5 decimal places. [2] (c) A student puts x = −1 into the expansion to estimate √3. Explain why this does not work. [2]

9. f(x) = (5 − x)/((1 + x)(1 − 2x)). (a) Write f(x) as a sum of partial fractions. [3] (b) Use this to write f(x) as a series up to and including the term in x³. [3] (c) Give the values of x for which your series holds. [2]

10. g(x) = (6 − 3x)/((1 + x)(1 − 2x)²). (a) Find constants A, B and C such that g(x) = A/(1 + x) + B/(1 − 2x) + C/(1 − 2x)². [4] (b) Use this to find the series for g(x) as far as the term in x³. [4] (c) Give the values of x for which this series holds. [1]

11. The expansion of (1 + ax)ⁿ begins 1 − 6x + 27x² + …, where a and n are constants and n is not a positive integer. (a) Find the values of n and a. [4] (b) Find the coefficient of x³ in the expansion. [2] (c) Give the values of x for which this series holds. [2]

Answers

1. nC2 = n(n − 1)/2 = 78, so n(n − 1) = 156 [1]. n² − n − 156 = 0, so (n − 13)(n + 12) = 0 [1]. n must be positive, so n = 13 [1]. [3] Examiner insight: Rejecting n = −12 with a reason shows you know n is a positive integer here. Quoting only 13 with no equation leaves nothing to credit if it is wrong.

2. Coefficients for n = 4 are 1, 4, 6, 4, 1 [1]. (3 − 4x)⁴ = 3⁴ + 4(3³)(−4x) + 6(3²)(−4x)² + 4(3)(−4x)³ + (−4x)⁴ [1] = 81 − 432x + 864x² − 768x³ + 256x⁴. First three terms simplified correctly [1], last two correctly [1]: 81 − 432x + 864x² − 768x³ + 256x⁴. [4] Examiner insight: Write the unsimplified line first. A slip such as (−4x)² = −16x² is then visible as one error, not a whole failed question. Alternate signs are a quick check.

3. (a) x²: 6C2 × 3⁴ × (−1)² = 15 × 81 = 1215 [1]. x³: 6C3 × 3³ × (−1)³ = 20 × 27 × (−1) = −540 [1]. [2] (b) The x³ terms come from 1 × (x³ term) and 2x × (x² term) [1]. 1 × (−540) + 2 × 1215 [1] = −540 + 2430 = 1890 [1]. [3] Examiner insight: Use your part (a) values. Expanding all seven terms of the second bracket first is slower and invites arithmetic slips. Show which pairs of terms you multiply.

4. Coefficient of x² is 6C2 × 2⁴ × k² [1] = 15 × 16 × k² = 240k² [1]. 240k² = 960 [1], so k² = 4 and k = 2 or k = −2 [1]. [4] Examiner insight: “Possible values” signals a square root. Giving only k = 2 drops the second solution.

5. (a) P(X = 4) = 6C4 × 0.8⁴ × 0.2² [1] = 15 × 0.4096 × 0.04 = 0.24576 = 0.246 [1]. [2] (b) P(X = 5) = 6C5 × 0.8⁵ × 0.2 = 6 × 0.32768 × 0.2 = 0.393216 [1]. P(X = 6) = 0.8⁶ = 0.262144 [1]. Sum = 0.65536 = 0.655 [1]. [3] (c) (0.2 + 0.8)⁶ = 1⁶ = 1, and its seven terms are the probabilities of 0, 1, …, 6 scores, so they add to 1 [1]. P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.2⁶ = 1 − 0.000064 = 0.999936 [1]. [2] Examiner insight: Part (c) asks for an explanation. Quote the sum of the terms and the value of the bracket. Leave the unrounded figure until the end, then round to 3 significant figures once.

6. (8 − 3x)^(−1/3) = 8^(−1/3)(1 − 3x/8)^(−1/3) = (1/2)(1 + u)^(−1/3), u = −3x/8 [1]. (1 + u)^(−1/3) = 1 − u/3 + 2u²/9 − 14u³/81 + … [1] Substituting: 1 + x/8 [1] + x²/32 [1] + 7x³/768 [1]. Multiply by 1/2: 1/2 + x/16 + x²/64 + 7x³/1536 + … Valid when |3x/8| < 1, so |x| < 8/3 [1]. [6] Examiner insight: Taking out 8^(−1/3) = 1/2 first, and then multiplying every term by it, is an easy step to drop. Check the sign: u is negative, so the x and x³ terms in (1 + u) come from −u/3 and −14u³/81 and end up positive.

7. (a) (3 − 2x)⁻² = 3⁻²(1 − 2x/3)⁻² [1], valid for |2x/3| < 1, so |x| < 3/2 [1]. [2] (b) (5 + 2x)^(1/2) = √5 (1 + 2x/5)^(1/2), so the condition is |2x/5| < 1 [1]. The correct range is |x| < 5/2; the student inverted the fraction [1]. [2] (c) The power is a positive integer, so the expansion is finite (seven terms) and holds for all x [1]. [1] Examiner insight: Divide the constant by the coefficient of x, not the other way round. A one-line check is whether the boundary makes bx/a equal to 1.

8. (a) (1 − 2x)^(1/2) = 1 + (1/2)(−2x) + (1/2)(−1/2)/2² + (1/2)(−1/2)(−3/2)/6³ [1]. = 1 − x [1] − x²/2 − x³/2 [1]. 1 − x − x²/2 − x³/2 + … [3] (b) 1 − 2x = 0.94 gives x = 0.03 [1]. 1 − 0.03 − 0.00045 − 0.0000135 = 0.9695365, so 0.96954 [1]. [2] (c) 1 − 2x = 3 needs x = −1 [1], which lies outside the validity range |x| < 1/2 [1], so the expansion cannot be used there. [2] Examiner insight: State the value of x you are substituting and check it against the validity range before using it. “Does not work” without naming the range condition is unlikely to earn credit.

9. (a) 5 − x = A(1 − 2x) + B(1 + x) [1]. x = −1: 6 = 3A, so A = 2. x = 1/2: 4.5 = 1.5B, so B = 3 [1]. f(x) = 2/(1 + x) + 3/(1 − 2x) [1]. [3] (b) 2(1 + x)⁻¹ = 2 − 2x + 2x² − 2x³ [1]. 3(1 − 2x)⁻¹ = 3 + 6x + 12x² + 24x³ [1]. Sum: 5 + 4x + 14x² + 22x³ + … [1]. [3] Check: the x⁰ term of f(0) = 5/1 = 5. (c) First part needs |x| < 1 [1]; second needs |2x| < 1, so |x| < 1/2. Both hold for |x| < 1/2 [1]. [2] Examiner insight: Take the smaller range, not the larger. Always pair a validity statement with the part of the expansion it belongs to.

10. (a) 6 − 3x = A(1 − 2x)² + B(1 + x)(1 − 2x) + C(1 + x) [1]. x = 1/2: 4.5 = 1.5C, so C = 3 [1]. x = −1: 9 = 9A, so A = 1 [1]. x = 0: 6 = A + B + C, so B = 2 [1]. [4] (b) (1 + x)⁻¹ = 1 − x + x² − x³ [1]. 2(1 − 2x)⁻¹ = 2 + 4x + 8x² + 16x³ [1]. 3(1 − 2x)⁻² = 3(1 + 4x + 12x² + 32x³) = 3 + 12x + 36x² + 96x³ [1]. Adding: 6 + 15x + 45x² + 111x³ + … [1]. [4] (c) The binding condition is |2x| < 1, so |x| < 1/2 [1]. [1] Examiner insight: Test your constants by putting x = 0 into the original and the split form. Both give 6, which catches most slips before the expansion starts.

11. (a) x: na = −6, so a = −6/n [1]. x²: n(n − 1)a²/2 = 27 [1]. Substituting: 36(n − 1)/(2n) = 27, so 18(n − 1) = 27n [1], giving n = −2 and a = 3 [1]. [4] (b) x³ coefficient = n(n − 1)(n − 2)a³/3! = (−2)(−3)(−4)(27)/6 [1] = −108 [1]. [2] (c) (1 + 3x)⁻² needs |3x| < 1 [1], so |x| < 1/3 [1]. [2] Examiner insight: Eliminate one unknown before solving; trying to solve both equations at once is where most time is lost. Check by expanding (1 + 3x)⁻² = 1 − 6x + 27x² − 108x³.

Where marks are usually lost

• Squaring only x in (bx)² and forgetting to square b.
• Not taking out aⁿ before expanding, or using aⁿ instead of a⁻ⁿ for a negative power.
• Losing the sign of u, which flips every odd-powered term.
• Giving k = 2 when k² = 4 also allows k = −2.
• Writing the validity as |x| < a/b upside down, or leaving it as |u| < 1.
• Taking the larger of two validity ranges after partial fractions.
• Using an x-value outside the range for an approximation.
• Not stating which pairs of terms combine in a product expansion.
• Rounding too early in a probability question.

Next steps

• Recap with the revision notes or relearn from the study guide.
• Tick off 1.04a to 1.04d on the checklist and see the course hub.
• Take one of the free 10-minute diagnostics.
• Mixed practice: Pure Mathematics practice.
• Book a free trial class.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR: sections 1.04a to 1.04d and 1.02y.