| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Use standard formulae to show result |
| Difficulty | Standard +0.3 Part (a) is a standard induction proof with straightforward algebra—expanding (k+2)(k+3)(k+4) and simplifying. Part (b) requires testing small values to find a counterexample (n=1 gives 24, n=2 gives 60, n=3 gives 120, n=4 gives 210—not divisible by 12). This is routine Further Maths content with no novel insight required, slightly above average due to the algebraic manipulation in the inductive step. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{n}(r+1)(r+2) = \sum_{r=1}^{n}(r^2 + 3r + 2) = \sum_{r=1}^{n}r^2 + \sum_{r=1}^{n}3r + \sum_{r=1}^{n}2\) | M1 | Splits into \(\sum ar^2 + \sum br + \sum c\); PI |
| \(S = \frac{n}{6}(n+1)(2n+1) + \frac{3n}{2}(n+1) + 2n\) | M1 | Substitutes for \(\sum r^2\) and \(\sum r\); allow one slip |
| \(\sum_{r=1}^{n}1 = n\) | B1 | States or uses; PI |
| \(= \frac{n}{2}(n+1)(2n+1) + \frac{9n}{2}(n+1) + 6(n+1)\) | M1 | Factorises out \((n+1)\); allow one slip |
| \((n+1)\left\{\frac{n}{2}(2n+1) + \frac{9n}{2} + 6\right\}\) | M1 | Simplifies to find second linear factor from quadratic; FT if all M1s awarded; allow one slip |
| \(= (n+1)(n^2 + 5n + 6)\) | ||
| \(= (n+1)(n+2)(n+3)\) | R1 | Completes rigorous argument; factorising must be clearly seen and all previous marks obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(n=4\): \(6 + 3\sum_{r=1}^{n}(r+1)(r+2) = (5)(6)(7) = 210\) | E1 | Chooses a multiple of 4 for \(n\); correct numerical value/expression |
| \(210\) is not a multiple of \(12\), so Alex's statement is false | E1 | Clear argument with concluding statement |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n}(r+1)(r+2) = \sum_{r=1}^{n}(r^2 + 3r + 2) = \sum_{r=1}^{n}r^2 + \sum_{r=1}^{n}3r + \sum_{r=1}^{n}2$ | M1 | Splits into $\sum ar^2 + \sum br + \sum c$; PI |
| $S = \frac{n}{6}(n+1)(2n+1) + \frac{3n}{2}(n+1) + 2n$ | M1 | Substitutes for $\sum r^2$ and $\sum r$; allow one slip |
| $\sum_{r=1}^{n}1 = n$ | B1 | States or uses; PI |
| $= \frac{n}{2}(n+1)(2n+1) + \frac{9n}{2}(n+1) + 6(n+1)$ | M1 | Factorises out $(n+1)$; allow one slip |
| $(n+1)\left\{\frac{n}{2}(2n+1) + \frac{9n}{2} + 6\right\}$ | M1 | Simplifies to find second linear factor from quadratic; FT if all M1s awarded; allow one slip |
| $= (n+1)(n^2 + 5n + 6)$ | | |
| $= (n+1)(n+2)(n+3)$ | R1 | Completes rigorous argument; factorising must be clearly seen and all previous marks obtained |
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=4$: $6 + 3\sum_{r=1}^{n}(r+1)(r+2) = (5)(6)(7) = 210$ | E1 | Chooses a multiple of 4 for $n$; correct numerical value/expression |
| $210$ is not a multiple of $12$, so Alex's statement is false | E1 | Clear argument with concluding statement |
---10
\begin{enumerate}[label=(\alph*)]
\item Prove that
$$6 + 3 \sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 2 ) = ( n + 1 ) ( n + 2 ) ( n + 3 )$$
[6 marks]\\
10
\item Alex substituted a few values of $n$ into the expression $( n + 1 ) ( n + 2 ) ( n + 3 )$ and made the statement:\\
"For all positive integers n,
$$6 + 3 \sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 2 )$$
is divisible by $12 . "$
Disprove Alex's statement.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 Q10 [8]}}