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0606/22 Feb/Mar 2026, Question 11: The Binomial Expansion

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 7 marks

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The question, in short

Expand (x² + 1/(2x))⁹ up to three terms in decreasing powers of x, then use the general expansion with positive integer n to prove a divisibility condition.

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Part (a) · 3 marks

Write the first three terms, in descending powers of x, of the binomial expansion of (x² + 1/(2x))⁹, simplifying each term.

  1. 1First term: (x²)⁹ = x¹⁸
  2. 2Second term: 9(x²)⁸(1/(2x)) = (9/2)x¹⁵ = 4.5x¹⁵
  3. 3Third term: ⁹C₂(x²)⁷(1/(2x))² = 36x¹⁴/(4x²) = 9x¹²

Answerx¹⁸ + 4.5x¹⁵ + 9x¹²

Part (b) · 4 marks

For (x² + 1/(2x))ⁿ with n a positive integer, given one term has no x, prove that n is a multiple of 3.

  1. 1General term: ⁿCᵣ (x²)^(n−r) (1/(2x))^r, power of x is 2(n − r) − r
  2. 2Independent of x: 2(n − r) = r
  3. 33r = 2n, so r = 2n/3
  4. 4r is an integer and 2 is not divisible by 3, so 3 divides n

AnswerIndependent term needs 2(n − r) = r, so n = 3r/2 and r = 2n/3; since r is an integer, n is divisible by 3 (shown)

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Also searched as: 0606/22/F/M/26 · 0606_m26_qp_22 · 0606 m26 22 · question 11.
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