November 2021 Paper 1 Q10
10 The first four terms of a quadratic sequence are
\[0 \qquad 1 \qquad 0 \qquad -3\]Work out an expression for the \(n\)th term. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (Second differences =) \(-2\) or \(-n^2\) | M1 | second differences seen at least once and not contradicted may be seen by the sequence |
| \(0 - -1 \quad 1 - -4 \quad 0 - -9 \quad (-3 - -16)\) or \(1 \quad 5 \quad 9 \quad (13)\) or \(-1 - 0 \quad -4 - 1 \quad -9 - 0 \quad (-16 - -3)\) or \(-1 \quad -5 \quad -9 \quad (-13)\) | M1dep | subtracts \(-n^2\) from the given terms or subtracts the given terms from \(-n^2\) |
| \(-n^2 + 4n - 3\) | A1 | oe eg \(4n - 3 - n^2\) |
| Alternative method 2 | ||
| Any three of \(a + b + c = 0\) \(4a + 2b + c = 1\) \(9a + 3b + c = 0\) \(16a + 4b + c = -3\) | M1 | using \(n\)th term \(= an^2 + bn + c\) |
| \(3a + b = 1\) and \(5a + b = -1\) or \(a = -1\) and \(b = 4\) | M1dep | oe obtains two equations in the same two variables |
| \(-n^2 + 4n - 3\) | A1 | oe eg \(4n - 3 - n^2\) |
| Alternative method 3 | ||
| (Second differences =) \(-2\) or \(-n^2\) | M1 | second differences seen at least once and not contradicted may be seen by the sequence |
| \(3a + b = 1\) and substitutes \(a = -1\) | M1dep | oe eg \(-3 + b = 1\) or \(b = 4\) |
| \(-n^2 + 4n - 3\) | A1 | oe eg \(4n - 3 - n^2\) |
Additional guidance
| Condone use of U\(_n\) | M2A1 |
| Condone working in different variable(s) eg \(-n^2 + 4x - 3\) | M2A1 |
| Answer \(-n^2 \ldots\) scores at least M1 | |
| Condone \(-n^2 + 4n - 3 = 0\) or \(n = -n^2 + 4n - 3\) | M2A1 |