0606/21 May/Jun 2026, Question 7: The Binomial Expansion
Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 6 marks
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The question, in short
Find the constant term, free of x, in the product expansion of (2 − 3x)⁶ and (1 + 1/(4x))².
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Solution · 6 marks
- 1(2 − 3x)⁶ = 64 − 576x + 2160x² + …
- 2(1 + 1/(4x))² = 1 + 1/(2x) + 1/(16x²)
- 3Term independent of x = 64 × 1 + (−576x)(1/(2x)) + (2160x²)(1/(16x²))
- 4= 64 − 288 + 135 = −89
Answer−89
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Also searched as: 0606/21/M/J/26 · 0606_s26_qp_21 · 0606 s26 21 · question 7.
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