S6 Maths: Binomial Theorem, Paper 4 — coefficients, constant terms and products
Paper 4 drills coefficient arithmetic from expansions to constant terms and products of two brackets, always via the general term.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Maths: Binomial Theorem |
| Level | S6 |
| # | Answer | One-line note |
|---|---|---|
| 1 | 27 + 27x + 9x² + x³ | Pair powers of 3 with powers of x: 27, 27, 9 and finally 1. |
| 2 | 20 | The coefficient is 6 choose 3, which is 20, since powers of 1 change nothing. |
| 3 | 54 | Take 4 choose 2 times minus 3 squared, so 6 times 9 gives 54. |
| 4 | 90x³ | The third term takes r equal to 2, so 5 choose 2 times 9 gives 90 times x cubed. |
| 5 | 720 | It is 5 choose 2 times 2 cubed times 3 squared, so 10 times 8 times 9 gives 720. |
| 6 | 28 | The power is 24 minus 4r, which vanishes at r equal to 6, and 8 choose 6 is 28. |
| 7 | 280 | It is 7 choose 3 times 2 cubed, so 35 times 8 gives 280. |
| 8 | n = 12 | Setting n(n minus 1) over 2 to 66 gives (n minus 12)(n plus 11), so n is 12. |
| 9 | 6 | The product is (1 minus x squared) to the fourth, so the x fourth power coefficient is 4 choose 2, which is 6. |
| 10 | 10 | The coefficient is 5 choose 2, which is 10; the 5 is only the power. |
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