4MA0 Higher June 2014 Paper 4R Q3
3 Here is a sequence of patterns made from centimetre squares.
(a) Find an expression, in terms of \(n\), for the total number of centimetre squares in Pattern number \(n\). (2)
A pattern in this sequence has 88 centimetre squares.
(b) Work out the Pattern number of this pattern. (2)
| Scheme | Marks |
|---|---|
| M1 | |
| \(3n + 1\) oe | A1 |
| (2) |
Notes
M1: for \(3n + k\) (\(k\) may be zero) oe
A1: need not be simplified eg. \(4 + 3(n - 1)\)
NB: \(n = 3n + 1\) gains M1 A0
| Scheme | Marks |
|---|---|
| \(3n + 1 = 88\) or \((88 - 1) \div 3\) | M1 |
| 29 | A1 |
| (2) | |
| (4 marks) |
Notes
M1: ft “\(3n + 1\)” = 88
NB. Only ft if their expression is of the form \(an + b\) where \(a > 1\) and \(b \ne 0\)
A1: ft NB. unrounded answer must be an integer