| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Standard summation formulae application |
| Difficulty | Standard +0.3 Part (a) requires straightforward application of standard summation formulae (∑r³ and ∑r) and algebraic manipulation to reach the required form—routine for FP1 students. Part (b) involves substituting n=1000 into the factorised form and recognising that 2008 is a factor, which requires some insight but is a standard 'show that' extension. Overall slightly easier than average A-level difficulty. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Use of formula for \(\sum r^3\) or \(\sum r\) | M1 | |
| \(n\) is a factor of the expression | m1 | clearly shown |
| So is \((n+1)\) | m1 | ditto |
| \(S_n = \frac{1}{4}n(n+1)(n^2+n-12)\) | A1 | |
| \(= \frac{1}{4}n(n+1)(n+4)(n-3)\) | A1F | ft wrong value for \(k\) |
| Subtotal | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(n = 1000\) substituted into expression | m1 | The factor 1004, or \(1000+4\), seen |
| Conclusion convincingly shown | A1 | not '\(2008 \times 124749625\)' |
| Need \(\frac{1000}{4}\) is even, hence conclusion | OE | |
| Subtotal | 2 | |
| Total | 7 |
## Question 4(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Use of formula for $\sum r^3$ or $\sum r$ | M1 | |
| $n$ is a factor of the expression | m1 | clearly shown |
| So is $(n+1)$ | m1 | ditto |
| $S_n = \frac{1}{4}n(n+1)(n^2+n-12)$ | A1 | |
| $= \frac{1}{4}n(n+1)(n+4)(n-3)$ | A1F | ft wrong value for $k$ |
| **Subtotal** | **5** | |
## Question 4(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $n = 1000$ substituted into expression | m1 | The factor 1004, or $1000+4$, seen |
| Conclusion convincingly shown | A1 | not '$2008 \times 124749625$' |
| Need $\frac{1000}{4}$ is even, hence conclusion | | OE |
| **Subtotal** | **2** | |
| **Total** | **7** | |
---4
\begin{enumerate}[label=(\alph*)]
\item Find
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } - 6 r \right)$$
expressing your answer in the form
$$k n ( n + 1 ) ( n + p ) ( n + q )$$
where $k$ is a fraction and $p$ and $q$ are integers.
\item It is given that
$$S = \sum _ { r = 1 } ^ { 1000 } \left( r ^ { 3 } - 6 r \right)$$
Without calculating the value of $S$, show that $S$ is a multiple of 2008 .
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2008 Q4 [7]}}