Higher June 2025 Paper 1 Q10
10 Prove that the difference in the squares of two consecutive even numbers is always a multiple of 4 (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | M1 | for writing expressions for two consecutive even numbers, eg \(2n\) and \(2n + 2\) or \(2n - 2\) and \(2n\) or \(2n + 2\) and \(2n + 4\) oe (assuming \(n\) is any integer) |
| M1 | (dep M1) for correctly expanding the squares of both expressions, eg \((2n + 2)^2 = 4n^2 + 4n + 4n + 4\) and \((2n)^2 = 4n^2\) or \((2n)^2 = 4n^2\) and \((2n - 2)^2 = 4n^2 - 4n - 4n + 4\) or \((2n + 4)^2 = 4n^2 + 8n + 8n + 16\) and \((2n + 2)^2 = 4n^2 + 4n + 4n + 4\) or for a correct expression using the difference of two squares eg \((2n + 2 + 2n)(2n + 2 - 2n)\) oe or \((2n + 2n - 2)(2n - (2n - 2))\) oe | |
| C1 | for a complete proof without any errors leading to eg \(4(2n + 1)\) or \(4(2n - 1)\) or to a statement that eg \(8n + 4\) is a multiple of 4 because \(8n\) and 4 are both multiples of 4 |
Additional guidance
For both M marks accept use of linear expressions with a difference of 2,
eg \(n\) and \(n + 2\) or \(n + 1\) and \(n + 3\)
Expressions need not be simplified for this mark
\(4n^2 + 8n + 4 - 4n^2 = 8n + 4\)
\(4n^2 - (4n^2 - 8n + 4) = 8n - 4\)
Accept eg \[\begin{aligned}(2n)^2 - (2n + 2)^2 &= 4n^2 - (4n^2 + 4n + 4n + 4)\\ &= -8n - 4\\ &= 4(-2n - 1)\end{aligned}\]
A proof using eg \(n\) and \(n + 2\) must include a statement that \(n\) is even for the C mark to be awarded and a proof using eg \(n + 1\) and \(n + 3\) must include a statement that \(n\) is odd