| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Finding n for given sum value |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring partial fractions, method of differences (a non-trivial telescoping series technique), and solving an inequality involving the derived formula. While each component is standard FP2 material, the combination and the final part requiring algebraic manipulation to find n pushes this above average difficulty for A-level, though it remains a fairly typical Further Maths question rather than requiring exceptional insight. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(A = \frac{1}{2}, \quad B = -\frac{1}{2}\) | B1, B1F | 2 marks |
| (b) Method of differences clearly shown | M1 | |
| \(\text{Sum} = \frac{1}{2}\left(1 - \frac{1}{2n+1}\right)\) | A1 | |
| \(= \frac{n}{2n+1}\) | A1 | 3 marks |
| (c) \(\frac{1}{2(2n+1)} < 0.001\) or \(\frac{n}{2n+1} > 0.499\) | M1 | Condone use of equals sign |
| \(1 < 0.004n + 0.002\) or \(n > 0.998n + 0.499\) | A1 | |
| \(n > \frac{0.998}{0.004}\) or \(0.004n > 0.998\) | A1 | 3 marks |
| \(n = 250\) | A1F |
**(a)** $A = \frac{1}{2}, \quad B = -\frac{1}{2}$ | B1, B1F | 2 marks | For either $A$ or $B$; For the other
**(b)** Method of differences clearly shown | M1 |
$\text{Sum} = \frac{1}{2}\left(1 - \frac{1}{2n+1}\right)$ | A1 |
$= \frac{n}{2n+1}$ | A1 | 3 marks | AG
**(c)** $\frac{1}{2(2n+1)} < 0.001$ or $\frac{n}{2n+1} > 0.499$ | M1 | Condone use of equals sign
$1 < 0.004n + 0.002$ or $n > 0.998n + 0.499$ | A1 |
$n > \frac{0.998}{0.004}$ or $0.004n > 0.998$ | A1 | 3 marks | OE; ft if say 0.4999 used. If method of trial and improvement used, award full marks for completely correct solution showing working
$n = 250$ | A1F |
**Total: 8 marks**2
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\frac { 1 } { 4 r ^ { 2 } - 1 } = \frac { A } { 2 r - 1 } + \frac { B } { 2 r + 1 }$$
find the values of $A$ and $B$.
\item Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }$$
\item Find the least value of $n$ for which $\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 }$ differs from 0.5 by less than 0.001 .
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q2 [8]}}