Standard +0.3 This is a straightforward divisibility proof requiring students to substitute n=2k and expand to show n³+4=8k³+4=4(2k³+1). Recognizing that 2k³+1 is odd completes the proof. While it requires algebraic manipulation and understanding of even/odd properties, it's a standard AS-level proof technique with clear structure and no novel insight needed.
\(n = 2k\) [where \(k\) is an integer], \(n^3 + 4 = (2k)^3 + 4 = 4(2k^3 + 1)\)
M1
Uses the evenness of \(n\) in algebraic form. Allow if they state that the cube of an even number is a multiple of 8
Which is a multiple of 4
A1
Also allow for a factorised expression, or division by 4. Also allow an argument that each term of the sum is a multiple of 4
But \((2k^3 + 1)\) is always odd so not a multiple of 8
A1 [3]
Any clear argument leading to the statement that the expression is not a multiple of 8 www (See appendix for exemplars)
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n = 2k$ [where $k$ is an integer], $n^3 + 4 = (2k)^3 + 4 = 4(2k^3 + 1)$ | M1 | Uses the evenness of $n$ in algebraic form. Allow if they state that the cube of an even number is a multiple of 8 |
| Which is a multiple of 4 | A1 | Also allow for a factorised expression, or division by 4. Also allow an argument that each term of the sum is a multiple of 4 |
| But $(2k^3 + 1)$ is always odd **so not a multiple of 8** | A1 [3] | Any clear argument leading to the statement that the expression is not a multiple of 8 www (See appendix for exemplars) |
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3 Prove that, when $n$ is an even number, $n ^ { 3 } + 4$ is a multiple of 4 but not a multiple of 8 .
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q3 [3]}}