| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| 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 | Challenging +1.2 This is a Further Maths question requiring method of differences with a non-standard summation range (n to 2n). Part (a) is routine algebraic verification, but part (b) requires careful handling of telescoping sums with shifted limits and subsequent algebraic manipulation to reach the given form—more demanding than standard A-level but a familiar FP2 technique. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| - \(f(r) - f(r-1)\) | ||
| - \(= \frac{1}{4}r^2(r+1)^2 - \frac{1}{4}(r-1)^2r^2\) | M1 | |
| - \(= \frac{1}{4}r^2\left(r^2 + 2r + 1 - r^2 + 2r - 1\right)\) | A1 | Correct expansions of \((r+1)^2\) and \((r-1)^2\) |
| - \(= r^3\) | A1 | 3 marks; AG |
| Answer | Marks | Guidance |
|---|---|---|
| - \(r = n: n^3 = \frac{1}{4}n^2(n+1)^2 - \frac{1}{4}(n-1)^2n^2\) | M1, A1 | For either \(r = n\) or \(r = 2n\); PI |
| - \(r = 2n:\) | ||
| - \((2n)^3 = \frac{1}{4}(2n)^2(2n+1)^2 - \frac{1}{4}(2n-1)^2(2n)^2\) | A1 | |
| - \(\sum_{r=n}^{2n} r^3 = \frac{1}{4}4n^2(2n+1)^2 - \frac{1}{4}(n-1)^2n^2\) | M1 | |
| - \(= \frac{3}{4}n^2(5n+1)(n+1)\) | A1 | 5 marks; AG |
**(a)**
- $f(r) - f(r-1)$ | |
- $= \frac{1}{4}r^2(r+1)^2 - \frac{1}{4}(r-1)^2r^2$ | M1 |
- $= \frac{1}{4}r^2\left(r^2 + 2r + 1 - r^2 + 2r - 1\right)$ | A1 | Correct expansions of $(r+1)^2$ and $(r-1)^2$
- $= r^3$ | A1 | 3 marks; AG
- **Total: 3 marks**
**(b)**
- $r = n: n^3 = \frac{1}{4}n^2(n+1)^2 - \frac{1}{4}(n-1)^2n^2$ | M1, A1 | For either $r = n$ or $r = 2n$; PI
- $r = 2n:$ | |
- $(2n)^3 = \frac{1}{4}(2n)^2(2n+1)^2 - \frac{1}{4}(2n-1)^2(2n)^2$ | A1 |
- $\sum_{r=n}^{2n} r^3 = \frac{1}{4}4n^2(2n+1)^2 - \frac{1}{4}(n-1)^2n^2$ | M1 |
- $= \frac{3}{4}n^2(5n+1)(n+1)$ | A1 | 5 marks; AG
- Alternatively: $\sum_{r=n}^{2n} r^3$ and $\sum_{r=1}^{n-1} r^3$ stated M1A1A1 (M1 for either); Difference M1; Answer A1
- **Total: 5 marks**
**Question 3 Total: 8 marks**
---3
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ( r ) = \frac { 1 } { 4 } r ^ { 2 } ( r + 1 ) ^ { 2 }$, show that
$$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = r ^ { 3 }$$
\item Use the method of differences to show that
$$\sum _ { r = n } ^ { 2 n } r ^ { 3 } = \frac { 3 } { 4 } n ^ { 2 } ( n + 1 ) ( 5 n + 1 )$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q3 [8]}}