Higher November 2023 Paper 2 Q9
9 Here is the term-to-term rule for a sequence.
| Double the previous term and add 3 |
The first three terms of the sequence are \(\quad a + 1 \quad 2a + 5 \quad 4a + 13\)
Show that the sum of the first four terms is a multiple of 3 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(8a + 29\) | B1 | oe eg \(2(4a + 13) + 3\) |
| \(15a + 48\) | B1ft | correct or ft B0 only their \(8a + 29\) must be in the form \(na + c\) where \(n \ne 0\) and \(c \ne 0\) implied by \(3(5a + 16)\) |
| \(3(5a + 16)\) or \(15 = 5 \times 3\) and \(48 = 16 \times 3\) | B1 | oe eg \(5a + 16\) so it divides by 3 |
Additional guidance
| Ignore use of substitution as an attempt to show divisibility | |
| Ignore further non-contradictory statements | |
| Further simplification eg \(15a + 48 = 63\) which is \(21 \times 3\) | B1B1B0 |
| For the 1st B1 accept \(8a + 29\) embedded in a calculation for the sum of the first four terms eg \(a + 1 + 2a + 5 + 4a + 13 + 8a + 29\) | |
| For the 2nd B1 accept \(15a + 48\) embedded in a calculation to show divisibility eg \(\dfrac{15a + 48}{3} = 5a + 16\) | |
| For the 3rd B1 accept 15 is a multiple of 3 and 48 is a multiple of 3 | |
| \(8a + 29\) \(a + 2a + 4a + 8a = 15a \quad 1 + 5 + 13 + 29 = 48\) but \(15a + 48\) not seen \(15 = 5 \times 3\) and \(48 = 16 \times 3\) | B1 B0 B1 |