Higher June 2023 Paper 2 Q19
19 Here are the first four terms of a quadratic sequence.
\[3 \qquad 20 \qquad 47 \qquad 84\]Work out an expression for the \(n\)th term of the sequence. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 \(n\)th term \(= an^2 + bn + c\) | ||
| (second differences =) 10 or \(a = 5\) or \(5n^2\) | M1 | second difference seen at least once and not contradicted by a different value unless recovered may be seen by the sequence |
| \(3 - 5 \times 1^2\) and \(20 - 5 \times 2^2\) or \(-2\) and 0 or \(b = 2\) or \(2n\) | M1dep | oe subtraction of \(5n^2\) from any two consecutive terms eg \(47 - 5 \times 3^2\) and \(84 - 5 \times 4^2\) or 2 and 4 implied by \(5n^2 + 2n \ldots\) |
| \(5 \times 1^2 + 2 \times 1 + c = 3\) or \(5 + 2 + c = 3\) or (\(2n + c\) and) \(2 \times 1 + c = -2\) | M1dep | oe substitution of \(a = 5\) and \(b = 2\) eg \(5 \times 2^2 + 2 \times 2 + c = 20\) or oe use of \(2n + c\) and another term eg (\(2n + c\) and) \(2 \times 2 + c = 0\) |
| \(5n^2 + 2n - 4\) | A1 | terms in any order SC2 \(a = 5\) and \(c = -4\) SC1 \(c = -4\) |
| Alternative method 2 \(n\)th term \(= an^2 + bn + c\) | ||
| (second differences =) 10 or \(a = 5\) or \(5n^2\) | M1 | second difference seen at least once and not contradicted by a different value unless recovered may be seen by the sequence |
| \(3 \times 5 + b = 17\) or \(b = 2\) or \(2n\) | M1dep | oe substitution of \(a = 5\) eg \(5 \times 5 + b = 27\) implied by \(5n^2 + 2n \ldots\) |
| \(5 \times 1^2 + 2 \times 1 + c = 3\) or \(5 + 2 + c = 3\) | M1dep | oe substitution of \(a = 5\) and \(b = 2\) eg \(5 \times 2^2 + 2 \times 2 + c = 20\) |
| \(5n^2 + 2n - 4\) | A1 | terms in any order SC2 \(a = 5\) and \(c = -4\) SC1 \(c = -4\) |
| Alternative method 3 \(n\)th term \(= an^2 + bn + c\) | ||
| Any 3 of \(a + b + c = 3\) \(4a + 2b + c = 20\) \(9a + 3b + c = 47\) \(16a + 4b + c = 84\) | M1 | oe 3 equations |
| \(3a + b = 17\) and \(5a + b = 27\) or \(a = 5\) and \(b = 2\) | M1dep | oe pair of equations in \(a\) and \(b\) eg \(8a + 2b = 44\) and \(15a + 3b = 81\) implied by \(5n^2 + 2n \ldots\) |
| \(5 \times 1^2 + 2 \times 1 + c = 3\) or \(5 + 2 + c = 3\) | M1dep | oe substitution of \(a = 5\) and \(b = 2\) eg \(5 \times 2^2 + 2 \times 2 + c = 20\) |
| \(5n^2 + 2n - 4\) | A1 | terms in any order SC2 \(a = 5\) and \(c = -4\) SC1 \(c = -4\) |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Second differences = 10 scores M1 even if used incorrectly eg \(10n \ldots\) | |
| Condone \(n = 5n^2 + 2n - 4\) or \(5n^2 + 2n - 4 = 0\) | M3A1 |
| Condone working in a different variable eg \(5x^2 + 2x - 4\) | M3A1 |
| The 3rd method mark cannot be implied ie \(c = -4\) is only awarded M3 if the previous two method marks are seen | |
| Alt 1 2nd M1 cannot be awarded for subtracting in the wrong order unless recovered | |
| SC2 or SC1 can be awarded from work seen in the working lines | |
| SC2 or SC1 can be implied by a quadratic answer eg1 answer \(5n^2 + 6n - 4\) eg2 answer \(10n^2 + 3n - 4\) | SC2 SC1 |