Higher June 2025 Paper 1R Q17
17 Prove that, for any three numbers which are consecutive multiples of 4, the difference between the square of the largest number and the square of the smallest number is always a multiple of 64
Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
| eg \(4n, 4n + 4, 4n + 8\) or \(4n, 4(n + 1), 4(n + 2)\) or \(4n - 4, 4n, 4n + 4\) or \(4(n - 1), 4n, 4(n + 1)\) | M1 |
| eg \((4n + 8)^2 - (4n)^2\) (= \(16n^2 + 64n + 64 - 16n^2\)) or \((4(n + 2))^2 - (4n)^2\) (= \(16n^2 + 64n + 64 - 16n^2\)) or \((4n + 4)^2 - (4n - 4)^2\) (= \(16n^2 + 32n + 16 - 16n^2 + 32n - 16\)) or \((4(n + 1))^2 - (4(n - 1))^2\) (= \(16n^2 + 32n + 16 - 16n^2 + 32n - 16\)) | M1 |
| eg \((4n + 8)^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2 = 64n + 64\) or \((4n + 8)^2 - (4n)^2 = (4n + 8 + 4n)(4n + 8 - 4n) = 8(8n + 8) = 64n + 64\) or \((4(n + 2))^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2 = 64n + 64\) or \((4(n + 2))^2 - (4n)^2 = (4(n + 2) + 4n)(4(n + 2) - 4n) = 8(8n + 8) = 64n + 64\) or \((4n + 4)^2 - (4n - 4)^2 = 16n^2 + 32n + 16 - 16n^2 + 32n - 16 = 64n\) or \((4n + 4)^2 - (4n - 4)^2 = (4n + 4 + 4n - 4)(4n + 4 - 4n + 4) = 8n \times 8 = 64n\) or \((4(n + 1))^2 - (4(n - 1))^2 = 16n^2 + 32n + 16 - 16n^2 + 32n - 16 = 64n\) or \((4(n + 1))^2 - (4(n - 1))^2 = (4(n + 1) + 4(n - 1))(4(n + 1) - 4(n - 1)) = 8n \times 8 = 64n\) Working required Answer: correctly shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for correct expressions for 3 consecutive multiples of 4 (any letter can be used) may just see the first and third multiple for this mark
M1: for squaring the largest and smallest multiple of 4 and subtracting (no need to expand or simplify for this mark)
A1: dep on M2, for use of algebra to show correct conclusion
In the last line of working, \(4(n - 1)\) is used twice (corrected from the printed mark scheme: “\(4(n - 4)\)”).