Higher November 2019 Paper 2 Q21
21 \(n\) is the middle integer of three consecutive positive integers.
The three integers are multiplied to give a product.
\(n\) is then added to the product.
Prove that the result is a cube number. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(n - 1\) and \(n\) and \(n + 1\) | M1 | oe eg \((n - 1)n(n + 1)\) or \(n(n - 1)(n + 1)\) |
| \(n(n^2 + n - n - 1)\) with M1 seen or \(n(n^2 - 1)\) with M1 seen or \((n^2 - n)(n + 1)\) with M1 seen or \((n^2 + n)(n - 1)\) with M1 seen | M1dep | |
| \(n^3 - n^2 + n^2 - n + n\) with M2 seen or \(n^3 - n + n\) with M2 seen | M1dep | |
| \(n^3\) with M3 seen | A1 | |
| Alternative method 2 | ||
| \(x\) and \(x + 1\) and \(x + 2\) | M1 | oe eg \(x(x + 1)(x + 2)\) or \((x + 1)x(x + 2)\) |
| \((x^2 + x)(x + 2)\) with M1 seen or \((x^2 + 2x)(x + 1)\) with M1 seen or \(x(x^2 + 2x + x + 2)\) with M1 seen or \(x(x^2 + 3x + 2)\) with M1 seen | M1dep | |
| \(x^3 + 3x^2 + 2x + x + 1\) with M2 seen or \(x^3 + x^2 + 2x^2 + 2x + x + 1\) with M2 seen | M1dep | |
| \(x^3 + 3x^2 + 3x + 1\) and \((x + 1)^3\) with M3 seen | A1 | allow \(x^3 + 3x^2 + 3x + 1\) and \(n^3\) with M3 seen if \(n = x + 1\) stated |
Additional guidance
| Only numerical example(s) | Zero |
| Condone use of any letter eg \(N\) |