| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Algebraic proof about integers |
| Difficulty | Moderate -0.3 This is a straightforward algebraic proof requiring factorization of n³-n = n(n-1)(n+1) and recognizing consecutive odd numbers give an even×odd×even product (divisible by 4). Part (b) requires only testing n=2 as a counterexample. Below average difficulty as it's a standard proof technique with clear structure and minimal steps. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Selects correct strategy: odd number is \(2k \pm 1\) | B1 | May be in any variable (condone use of \(n\)) |
| Attempts to simplify \((2k\pm1)^3 - (2k\pm1) = \ldots\) | M1 | Condone errors in expanding; either coefficient of \(k\) term or constant must change |
| Factorise \(8k^3 \pm 12k^2 \pm 4k = 4k(2k^2 \pm 3k \pm 1)\) | dM1 | Attempts to take factor of 4 or \(4k\) from cubic |
| Correct work with statement: \(4\times\ldots\) is a multiple of 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Any counter example with correct statement, e.g. \(2^3 - 2 = 6\) which is not a multiple of 4 | B1 | Any valid counter example with correct statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Factorise \(k^3 - k = k(k-1)(k+1)\) | B1 | |
| States if \(k\) is odd then both \(k-1\) and \(k+1\) are even | M1 | |
| States \(k-1\) multiplied by \(k+1\) is therefore a multiple of 4 | dM1 | |
| Concludes \(k^3 - k\) is a multiple of 4 as it is odd \(\times\) multiple of 4 | A1 |
## Question 10:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Selects correct strategy: odd number is $2k \pm 1$ | B1 | May be in any variable (condone use of $n$) |
| Attempts to simplify $(2k\pm1)^3 - (2k\pm1) = \ldots$ | M1 | Condone errors in expanding; either coefficient of $k$ term or constant must change |
| Factorise $8k^3 \pm 12k^2 \pm 4k = 4k(2k^2 \pm 3k \pm 1)$ | dM1 | Attempts to take factor of 4 or $4k$ from cubic |
| Correct work with statement: $4\times\ldots$ is a multiple of 4 | A1 | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Any counter example with correct statement, e.g. $2^3 - 2 = 6$ which is not a multiple of 4 | B1 | Any valid counter example with correct statement |
### Alt (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Factorise $k^3 - k = k(k-1)(k+1)$ | B1 | |
| States if $k$ is odd then both $k-1$ and $k+1$ are even | M1 | |
| States $k-1$ multiplied by $k+1$ is therefore a multiple of 4 | dM1 | |
| Concludes $k^3 - k$ is a multiple of 4 as it is odd $\times$ multiple of 4 | A1 | |
---\begin{enumerate}
\item A student is investigating the following statement about natural numbers.
\end{enumerate}
\begin{displayquote}
" $n ^ { 3 } - n$ is a multiple of 4 "\\
(a) Prove, using algebra, that the statement is true for all odd numbers.\\
(b) Use a counterexample to show that the statement is not always true.
\end{displayquote}
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q10 [5]}}