| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Standard +0.3 This is a straightforward application of standard summation formulae. Part (a) requires direct substitution of known formulae for Σr² and Σr, then algebraic simplification. Part (b) uses the result from (a) with the standard technique S₂ₙ - Sₙ to find the sum from n+1 to 2n. While it involves multiple steps, each step follows a routine procedure with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_n = 3\Sigma r^2 - 3\Sigma r + \Sigma 1\) | M1 | |
| Correct expressions substituted | m1 | At least for first two terms |
| Correct expansions | A1 | |
| \(\Sigma 1 = n\) | B1 | |
| Answer convincingly obtained | A1 | AG |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_{2n} - S_n\) attempted | M1 | Condone \(S_{2n}-S_{n+1}\) here |
| Answer \(7n^3\) | A1 | |
| Total | 7 |
## Question 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_n = 3\Sigma r^2 - 3\Sigma r + \Sigma 1$ | M1 | |
| Correct expressions substituted | m1 | At least for first two terms |
| Correct expansions | A1 | |
| $\Sigma 1 = n$ | B1 | |
| Answer convincingly obtained | A1 | AG |
| **Total** | **5** | |
## Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{2n} - S_n$ attempted | M1 | Condone $S_{2n}-S_{n+1}$ here |
| Answer $7n^3$ | A1 | |
| **Total** | **7** | |
---4 It is given that
$$S _ { n } = \sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } - 3 r + 1 \right)$$
\begin{enumerate}[label=(\alph*)]
\item Use the formulae for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that $S _ { n } = n ^ { 3 }$.
\item Hence show that $\sum _ { r = n + 1 } ^ { 2 n } \left( 3 r ^ { 2 } - 3 r + 1 \right) = k n ^ { 3 }$ for some integer $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2009 Q4 [7]}}