Specimen 2020 Paper 2 Q21
21 Show that \(\quad (2n + 3)^3 + n^3 \quad\) is divisible by 9 for all integer values of \(n\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4n^2 + 6n + 6n + 9\) or \(4n^2 + 12n + 9\) | M1 | allow one error implied by \(4n^2 + 12n + k\) or \(an^2 + 12n + 9\) |
| \(8n^3 + 12n^2 + 24n^2 + 36n + 18n + 27\) | M1dep | oe ft their \(4n^2 + 6n + 6n + 9\) allow one error |
| \(8n^3 + 36n^2 + 54n + 27\) or \(9n^3 + 36n^2 + 54n + 27\) | A1 | |
| \(9n^3 + 36n^2 + 54n + 27\) and \(9(n^3 + 4n^2 + 6n + 3)\) | A1 | oe eg \((9n^3 + 36n^2 + 54n + 27) \div 9\) \(= n^3 + 4n^2 + 6n + 3\) or \(9n^3 + 36n^2 + 54n + 27\) and all coefficients are divisible by 9 |