Higher June 2017 Paper 5 Q21
21 \(n\) is an integer.
(a) Explain why \(2n + 1\) is an odd number. [1]
(b) Prove that the difference between the squares of two consecutive odd numbers is a multiple of 8. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(2n\) is even and adding 1 gives an odd number oe | 1 | Must interpret the \(2n\) as even or not odd and then the +1 giving odd oe | Accept ‘\(2n\) is a multiple of 2’ for \(2n\) is even Accept 2 times any number is even oe for \(2n\) is even (as \(n\) is defined as an integer in the stem of question) Accept ‘next number’ or ‘odd’ for +1 Do not accept e.g. \(2n\) = even \(2n + 1\) is odd (does not interpret the 1) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((2n + 3)^2 - (2n + 1)^2\) oe | M2 | Allow \((2n + a)^2 - \{2n + (a \pm 2)\}^2\) where \(a\) is odd Or M1 for \(2n - 1\) or \(2n + 3\) used with \(2n + 1\) Allow \(\{2n + (a \pm 2)\}\) used with \((2n + a)\) where \(a\) is odd | Could use alternate correct expressions for consecutive odd numbers. Allow M and A marks if correct. Could reverse the algebraic terms their\((2n + 1)^2 - (2n + 3)^2\) leading to \(-8n - 8\), allow method and accuracy marks if correct. If brackets omitted allow recovery for M2 if correct expansion |
| \(4n^2 + 12n + 9 - 4n^2 - 4n - 1\) | M2 | Dep on M2 for expanding brackets in their expressions. Or M1indep for one correct expansion of their brackets | If seen alone and completely correct then implies previous M2 Allow \(4n^2 + 12n + 9 - (4n^2 + 4n + 1)\) |
| \(8n\) or \(8n + 8 = 8(n + 1)\) Or \(8n + 8\) is a multiple of 8 | A1 | With no errors or omissions seen. Correct for their two consecutive odd number expressions After 0 scored, Allow SC1 for two correctly evaluated numeric examples of subtracting consecutive odd squares isw | Accept \(-8n\) or \(-8n - 8\) oe if subtraction is reversed NB: M2M1A1 not possible – must earn all method marks for A mark e.g. \(7^2 - 5^2 = 24\) and \(3^2 - 5^2 = -16\) |