0606/22 May/Jun 2025, Question 10: The Binomial Expansion
Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 9 marks
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The question, in short
Given the first three terms 729x¹² + 972x¹⁰ + bx⁸ of the expansion of (3x² − a)ⁿ(1 + 1/x²)², find the constants a, b and n.
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Solution · 9 marks
- 1First term: (3x²)ⁿ = 729x¹², so 3ⁿ = 729 and n = 6
- 2(3x² − a)⁶ = 729x¹² − 1458ax¹⁰ + 1215a²x⁸ − …
- 3(1 + 1/x²)² = 1 + 2/x² + 1/x⁴
- 4x¹⁰: −1458a + 2 × 729 = 972, so a = 1/3
- 5x⁸: 1215a² − 2 × 1458a + 729 = b
- 6b = 135 − 972 + 729 = −108
Answera = 1/3, b = −108, n = 6
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