1MA0 Higher June 2012 Paper 2 Q21
21 Prove that
\[(2n + 3)^2 - (2n - 3)^2 \text{ is a multiple of 8}\]for all positive integer values of \(n\). (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(4n^2 + 12n + 3^2 - (4n^2 - 12n + 3^2)\) \(= 4n^2 + 12n + 9 - 4n^2 + 12n - 9\) \(= 24n\) \(= 8 \times 3n\) | Proof | 3 | M1 for 3 out of 4 terms correct in expansion of either \((2n + 3)^2\) or \((2n - 3)^2\) or \(((2n + 3) - (2n - 3))((2n + 3) + (2n - 3))\) A1 for \(24n\) from correct expansion of both brackets A1 (dep on A1) for \(24n\) is a multiple of 8 or \(24n = 8 \times 3n\) or \(24n \div 8 = 3n\) |