| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| 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 | Standard +0.8 This is a Further Maths question requiring understanding that u_r = S_r - S_{r-1}, algebraic manipulation to derive the general term, and then evaluating a sum from n+1 to 2n using the difference S_{2n} - S_n. The algebraic manipulation in part (b) is non-trivial but follows a standard technique. Moderately challenging for FP2 level. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_1 = S_1 = 1^2 \cdot 2 \cdot 3 = 6\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_2 = S_2 - S_1 = 42\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_n = S_n - S_{n-1}\) | M1 | |
| \(= n^2(n+1)(n+2) - (n-1)^2n(n+1)\) | A1 | |
| \(= n(n+1)(4n-1)\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=n+1}^{2n} u_r = S_{2n} - S_n\) | M1 | |
| \(= (2n)^2(2n+1)(2n+2) - n^2(n+1)(n+2)\) | A1 | |
| \(= 3n^2(n+1)(5n+2)\) | A1 | 3 marks |
### Part (a)(i)
$u_1 = S_1 = 1^2 \cdot 2 \cdot 3 = 6$ | B1 | 1 mark | AG
### Part (a)(ii)
$u_2 = S_2 - S_1 = 42$ | B1 | 1 mark | AG
### Part (a)(iii)
$u_n = S_n - S_{n-1}$ | M1 |
$= n^2(n+1)(n+2) - (n-1)^2n(n+1)$ | A1 |
$= n(n+1)(4n-1)$ | A1 | 3 marks | AG
### Part (b)
$\sum_{r=n+1}^{2n} u_r = S_{2n} - S_n$ | M1 |
$= (2n)^2(2n+1)(2n+2) - n^2(n+1)(n+2)$ | A1 |
$= 3n^2(n+1)(5n+2)$ | A1 | 3 marks | AG5 The sum to $r$ terms, $S _ { r }$, of a series is given by
$$S _ { r } = r ^ { 2 } ( r + 1 ) ( r + 2 )$$
Given that $u _ { r }$ is the $r$ th term of the series whose sum is $S _ { r }$, show that:\\
(a) (i) $u _ { 1 } = 6$;\\
(ii) $u _ { 2 } = 42$;\\
(iii) $\quad u _ { n } = n ( n + 1 ) ( 4 n - 1 )$.\\
(b) Show that
$$\sum _ { r = n + 1 } ^ { 2 n } u _ { r } = 3 n ^ { 2 } ( n + 1 ) ( 5 n + 2 )$$
\hfill \mbox{\textit{AQA FP2 2010 Q5 [8]}}