| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Algebraic proof about integers |
| Difficulty | Easy -1.2 This is a straightforward proof by cases (n even or n odd) requiring only basic algebraic manipulation. It's simpler than average A-level questions as it involves no calculus, no complex algebraic techniques, and the two-case structure is immediately apparent. This is a standard 'prove divisibility' exercise commonly seen in textbooks. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(n=2k\) or \(n=2k\pm1\) and attempts \(n(n^2+5)\) | M1 | For the key step attempting to find \(n(n^2+5)\) when \(n=2k\) or \(n=2k\pm1\) or equivalent representation. Condone use of e.g. \(n=2k+2\) or \(n=2n\pm1\). Condone use of e.g. \(n=2n\) and \(n=2n\pm1\) |
| Achieves \(2k(4k^2+5)\) (for \(n=2k\)) and states "even" or achieves \((2k+1)(4k^2+4k+6) = 2(2k+1)(2k^2+2k+3)\) (for \(n=2k+1\)) and states "even" or achieves \((2k-1)(4k^2-4k+6) = 2(2k-1)(2k^2-2k+3)\) (for \(n=2k-1\)) and states "even" | A1 | Achieves \(2k(4k^2+5)\) or e.g. \(2(4k^3+5k)\) and deduces this is even at the appropriate time. Or achieves \((2k\pm1)(4k^2\pm4k+6) = 2(2k\pm1)(2k^2\pm2k+3)\) and deduces this is even. Note that if the bracket is expanded to e.g. \(8k^3+12k^2+16k+6\) then stating "even" is insufficient — they would need to say e.g. even + even + even + even = even or equivalent. Note it is also acceptable to use a divisibility argument e.g. \(\frac{8k^3+10k}{2} = 4k^3+5k\) so \(8k^3+10k\) must be even. There should be no errors in the algebra but allow e.g. invisible brackets if they are "recovered". |
| Sets \(n=2k\) and \(n=2k\pm1\) and attempts \(n(n^2+5)\) | dM1 | Attempts \(n(n^2+5)\) when \(n=2k\) and \(n=2k\pm1\) or equivalent representation of odd and even. |
| Achieves \(2k(4k^2+5)\) (for \(n=2k\)) and achieves \((2k\pm1)(4k^2\pm4k+6) = 2(2k\pm1)(2k^2\pm2k+3)\) (for \(n=2k\pm1\)), states "even" for both, correct work with a final conclusion showing true for all \(n(\in\mathbb{N})\) or e.g. true for all even and odd numbers | A1 | Correct work and states even for both WITH a final conclusion. There should be no errors in the algebra but allow e.g. invisible brackets if they are "recovered". |
## Question 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $n=2k$ **or** $n=2k\pm1$ and attempts $n(n^2+5)$ | M1 | For the key step attempting to find $n(n^2+5)$ when $n=2k$ **or** $n=2k\pm1$ or equivalent representation. Condone use of e.g. $n=2k+2$ **or** $n=2n\pm1$. Condone use of e.g. $n=2n$ and $n=2n\pm1$ |
| Achieves $2k(4k^2+5)$ (for $n=2k$) and states "even" **or** achieves $(2k+1)(4k^2+4k+6) = 2(2k+1)(2k^2+2k+3)$ (for $n=2k+1$) and states "even" **or** achieves $(2k-1)(4k^2-4k+6) = 2(2k-1)(2k^2-2k+3)$ (for $n=2k-1$) and states "even" | A1 | Achieves $2k(4k^2+5)$ or e.g. $2(4k^3+5k)$ and deduces this is even at the appropriate time. Or achieves $(2k\pm1)(4k^2\pm4k+6) = 2(2k\pm1)(2k^2\pm2k+3)$ and deduces this is even. Note that if the bracket is expanded to e.g. $8k^3+12k^2+16k+6$ then stating "even" is insufficient — they would need to say e.g. even + even + even + even = even or equivalent. Note it is also acceptable to use a divisibility argument e.g. $\frac{8k^3+10k}{2} = 4k^3+5k$ so $8k^3+10k$ must be even. There should be no errors in the algebra but allow e.g. invisible brackets if they are "recovered". |
| Sets $n=2k$ **and** $n=2k\pm1$ and attempts $n(n^2+5)$ | dM1 | Attempts $n(n^2+5)$ when $n=2k$ **and** $n=2k\pm1$ or equivalent representation of odd and even. |
| Achieves $2k(4k^2+5)$ (for $n=2k$) **and** achieves $(2k\pm1)(4k^2\pm4k+6) = 2(2k\pm1)(2k^2\pm2k+3)$ (for $n=2k\pm1$), states "even" for both, correct work **with** a final conclusion showing true for all $n(\in\mathbb{N})$ or e.g. true for all even and odd numbers | A1 | Correct work and states even for both WITH a final conclusion. There should be no errors in the algebra but allow e.g. invisible brackets if they are "recovered". |
**(4 marks)**\begin{enumerate}
\item Prove, using algebra, that
\end{enumerate}
$$n \left( n ^ { 2 } + 5 \right)$$
is even for all $n \in \mathbb { N }$.
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q11 [4]}}