Higher November 2024 Paper 2 Q22
22 The \(n\)th term of a sequence is \(\quad n^2 - 30n + 236\)
By completing the square,
show that all the terms of the sequence have two or more digits. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((n - 15)^2\) | M1 | |
| \((n - 15)^2 - 15^2 + 236\) or \((n - 15)^2 - 225 + 236\) or \((n - 15)^2 + 11\) | M1dep | may be embedded in or implied by an inequality or equation eg \((n - 15)^2 - 15^2 + 236 = 10\) \((n - 15)^2 - 15^2 + 236 \gt 10\) \((n - 15)^2 \gt -1\) |
| Valid explanation with M1 seen | A1 | eg M1 seen and all the terms must be 11 or more or \((n - 15)^2 \geqslant 0\) and 11 is added |
Additional guidance
| Condone a different letter used eg \(x\) | |
| M2 and all the terms must be greater than 11 | M2A0 |
| M2 and the 15th term is the smallest | M2A0 |
| Least term is 11 with no working for completing the square | M0 |
| M2 and squaring a bracket always has two digits then adding 11 means it has at least two digits | M2A0 |
| \((n - 15)(n - 15)\) is equivalent to \((n - 15)^2\) | |
| \((n - 15n)^2\) | M0 |
| Ignore incorrect work after M2 eg \((n - 15)^2 + 11 = 0\) | M2 |
| Condone \((n - 15)^2\) is positive and 11 is added | M2A1 |