| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove by exhaustion (cases) |
| Difficulty | Standard +0.8 This is a Further Maths FP2 induction question requiring proof that f(n+1)-f(n) is divisible by 6, then using this in an induction proof. Part (a) is trivial, but part (b) requires algebraic manipulation, understanding divisibility arguments, and properly structuring an induction proof with the difference result. The combination of proving the difference property and then applying it in induction is more sophisticated than standard induction questions, placing it moderately above average difficulty. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) Clear reason given | E1 | 1 mark; Minimum O × E = E |
| 7(b)(i) \((k+1)[(k+1)^2 + 5] - k(k^2 + 5)\) | M1 | |
| \(= 3k^2 + 3k + 6\) | A1 | |
| \(k^2 + k = k(k+1) = M(2)\) | E1 | Must be shown |
| \(f(k+1) - f(k) = M(6)\) | E1 | 4 marks |
| 7(b)(ii) Assume true for \(n = k\) | M1 | Clear method |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore f(k+1) = M(6) + f(k) = M(6) + M(6) = M(6)\) | A1 | |
| True for \(n = 1\) | B1 | |
| \(P(n) \to P(n+1)\) and P(1) true | E1 | 4 marks (9 marks for question); Provided all other marks earned in (b)(ii) |
**7(a)** Clear reason given | E1 | 1 mark; Minimum O × E = E
**7(b)(i)** $(k+1)[(k+1)^2 + 5] - k(k^2 + 5)$ | M1
$= 3k^2 + 3k + 6$ | A1
$k^2 + k = k(k+1) = M(2)$ | E1 | Must be shown
$f(k+1) - f(k) = M(6)$ | E1 | 4 marks
**7(b)(ii)** Assume true for $n = k$ | M1 | Clear method
$f(k+1) - f(k) = M(6)$
$\therefore f(k+1) = M(6) + f(k) = M(6) + M(6) = M(6)$ | A1
True for $n = 1$ | B1
$P(n) \to P(n+1)$ and P(1) true | E1 | 4 marks (9 marks for question); Provided all other marks earned in (b)(ii)7
\begin{enumerate}[label=(\alph*)]
\item Explain why $n ( n + 1 )$ is a multiple of 2 when $n$ is an integer.
\item \begin{enumerate}[label=(\roman*)]
\item Given that
$$\mathrm { f } ( n ) = n \left( n ^ { 2 } + 5 \right)$$
show that $\mathrm { f } ( k + 1 ) - \mathrm { f } ( k )$, where $k$ is a positive integer, is a multiple of 6 .
\item Prove by induction that $\mathrm { f } ( n )$ is a multiple of 6 for all integers $n \geqslant 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q7 [9]}}