Higher January 2022 Paper 2 Q18
18 Prove that when the sum of the squares of any two consecutive odd numbers is divided by 8, the remainder is always 2
Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
| eg \((2n + 1)^2 + (2n - 1)^2\) or \((2n + 1)^2 + (2n + 3)^2\) oe | M1 |
| Eg \(4n^2 + 4n + 1 + 4n^2 - 4n + 1\) or \(8n^2 + 2\) or \(4n^2 + 4n + 1 + 4n^2 + 12n + 9\) or \(8n^2 + 16n + 10\) oe | M1 |
eg \(8 \times n^2 \underline{+\,2}\) \(\dfrac{8n^2 + 16n + 10}{8} = n^2 + 2n + \dfrac{10}{8}\) which shows a remainder of 2 or 10 – 8 = 2 or \(\dfrac{8n^2 + 16n + 10}{8} = n^2 + 2n + 1\) remainder 2 oe \(\dfrac{8n^2 + 16n + 10}{8} = n^2 + 2n + 1 + \dfrac{2}{8}\) oe \(8(n^2 + 2n + 1) + 2\) oe Answer: shown clearly | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for setting up a correct algebraic expression (any letter can be used) must have intention to add (may come after expanding)
M1: correct expansion of brackets and correct signs or a correct result.
A1: conclusion dep on M2 for eg \(8n^2 + 2\) and a suitable conclusion (may be shown as a calculation/in numbers). The conclusion must be an intention to show that the result is a multiple of 8 and there is 2 remaining.