Higher November 2024 Paper 5 Q22
22 A sequence has \(n\)th term \(2n^2 + 1\).
Prove algebraically that the sum of any two consecutive terms in this sequence is always a multiple of 4. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(2n^2 + 1 + 2(n + 1)^2 + 1\) | M2 | M1 for \(2(n + 1)^2 + 1\) or \(2(n - 1)^2 + 1\) | Accept \(2(n + a)^2 + 1 + 2(n + b)^2 + 1\) oe where there is a difference of 1 between \(a\) and \(b\) |
| \(2n^2 + 1 + 2n^2 + 4n + 2 + 1\) or better | M2 | Dep on M2 FT their consecutive algebraic terms For all brackets correctly expanded M1 for a squared bracket correctly expanded e.g. [2] \((n^2 + n + n + 1)\) or better | M1 accept any bracket squared expansion seen e.g. \((2n + 3)^2 = 4n^2 + 6n + 6n + 9\) or better or [2]\((n^2 - n - n + 1)\) if \(2(n - 1)^2\) used |
| \(4n^2 + 4n + 4\) | A1 | FT their consecutive algebraic terms after M2M2 earned | Accept e.g. \(4n^2 - 4n + 4\) if \(n\) and \(n - 1\) used |
| Correct conclusion e.g. \(4(n^2 + n + 1)\) and multiple of 4 Each of the terms is divisible by 4 so multiple of 4 | A1 | With no errors or omissions seen | |