June 2023 Paper 1 Q5
5 \(X_n\) and \(Y_n\) are the \(n\)th terms of two sequences.
\[X_n = (n - 1)(n + 1)\]\[Y_n = (n + 1)(n + 2)\]Prove that every term of the sequence with \(n\)th term \(\quad Y_n - X_n \quad\) is a multiple of 3 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((n + 1)(.........)\) or \(n^2 + n - n - 1\) or \(n^2 - 1\) or \(n^2 + 2n + n + 2\) or \(n^2 + 3n + 2\) | M1 | ie common factor of \((n + 1)\) taken out and attempt to subtract oe |
| \((n + 1)(n + 2 - (n - 1))\) or \(n^2 + 2n + n + 2 - (n^2 - n + n - 1)\) | M1dep | oe eg could be subtracted vertically |
| \((n + 1)(n + 2 - n + 1) = 3(n + 1)\) or \(3n + 3\) or \(n^2 + 3n + 2 - n^2 + 1 = 3n + 3\) or \(3(n + 1)\) | A1 | no statement needed at end of proof |
Additional guidance
| \(X_n - Y_n\) can be awarded method marks but A0 unless rescued at the end eg recovery \(-3(n + 1)\) so \(Y_n - X_n\) is a multiple of 3 | M2 A0 M2 A1 |
| Missing brackets can be recovered later in the proof (but if not recovered then second M should not be awarded). Must see \(3n + 3\) to show recovery eg \(n^2 + 3n + 3 - n^2 - 1\) (error in \(X_n\) but \(Y_n\) correct with missing bracket) | M2 A1 M1 M0 A0 |
| Condone use of \(x\) (or other letter) instead of \(n\). A mixture of letters would only gain M1. However \(Y_n - X_n = 3(n + 1) = 3x\) would be acceptable to show a multiple of 3 |