| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.8 This is a standard Further Maths method of differences question requiring algebraic expansion, telescoping series manipulation, and careful algebraic rearrangement to derive a well-known formula. While the technique is taught explicitly in FP2, executing it correctly requires multiple careful steps and the insight to connect the telescoping sum to the desired result, making it moderately challenging but within the expected scope of the syllabus. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Attempt to expand \((2r+1)^3 - (2r-1)^3\) | M1 | |
| \((2r+1)^3\) or \((2r-1)^3\) expanded | A1 | |
| \(24r^2 + 2\) | A1 | 3 marks; AG |
| (b) \(r = 1\): \(3^3 - 1^3 = 24 \times 1^2 + 2\) | M1A1 | 3 rows seen; Do not allow M1 for \((2n+1)^3 - 1\) not equal to anything |
| \(r = 2\): \(5^3 - 3^3 = 24 \times 2^2 + 2\) | ||
| \(r = n\): \((2n+1)^3 - (2n-1)^3 = 24 \times n^2 + 2\) | ||
| \((2n+1)^3 - 1 = 24\sum_{r=1}^{n} r^2 + 2n\) | A1 | |
| \(8n^3 + 12n^2 + 6n + 1 - 1 - 2n = 24\sum_{r=1}^{n} r^2\) | M1 | M1 for multiplication of bracket or taking \((2n+1)\) out as a factor |
| \(8n^3 + 12n^2 + 4n = 24\sum_{r=1}^{n} r^2\) | A1 | CAO |
| \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) | A1 | 6 marks; AG |
**(a)** Attempt to expand $(2r+1)^3 - (2r-1)^3$ | M1 |
$(2r+1)^3$ or $(2r-1)^3$ expanded | A1 |
$24r^2 + 2$ | A1 | 3 marks; AG
**(b)** $r = 1$: $3^3 - 1^3 = 24 \times 1^2 + 2$ | M1A1 | 3 rows seen; Do not allow M1 for $(2n+1)^3 - 1$ not equal to anything
$r = 2$: $5^3 - 3^3 = 24 \times 2^2 + 2$ | |
$r = n$: $(2n+1)^3 - (2n-1)^3 = 24 \times n^2 + 2$ | |
$(2n+1)^3 - 1 = 24\sum_{r=1}^{n} r^2 + 2n$ | A1 |
$8n^3 + 12n^2 + 6n + 1 - 1 - 2n = 24\sum_{r=1}^{n} r^2$ | M1 | M1 for multiplication of bracket or taking $(2n+1)$ out as a factor
$8n^3 + 12n^2 + 4n = 24\sum_{r=1}^{n} r^2$ | A1 | CAO
$\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ | A1 | 6 marks; AG
**Question 2 Total: 9 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item Show that
$$( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 } = 24 r ^ { 2 } + 2$$
\item Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q2 [9]}}