| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove polynomial divisibility property |
| Difficulty | Moderate -0.3 Part (i) is a routine factorisation requiring recognition of common factor n and difference of squares. Part (ii) requires showing the factorised form contains consecutive integers (hence divisible by 2 and 3), which is a standard proof technique taught in C3. The question is slightly easier than average as it's well-structured with a helpful hint, though it does require some mathematical reasoning beyond pure recall. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | n3 n = n(n2 1) |
| = n(n 1)(n + 1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | two correct factors | |
| 3 | (ii) | n 1, n and n + 1 are consecutive integers |
| Answer | Marks |
|---|---|
| [ n3 n is div by 6] | B1 |
Question 3:
3 | (i) | n3 n = n(n2 1)
= n(n 1)(n + 1) | B1
B1
[2] | two correct factors
3 | (ii) | n 1, n and n + 1 are consecutive integers
so at least one is even, and one is div by 3
[ n3 n is div by 6] | B1
B1
[2]\begin{enumerate}[label=(\roman*)]
\item Factorise fully $n^3 - n$. [2]
\item Hence prove that, if $n$ is an integer, $n^3 - n$ is divisible by 6. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q3 [4]}}