| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a standard FP2 method of differences question with straightforward algebra. Part (a) requires simple algebraic manipulation to find A=4, and part (b) is a textbook application of the telescoping series technique. While it's a Further Maths topic, the execution is routine with no conceptual challenges or novel insights required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(r) - f(r+1) = \frac{1}{4r-1} - \frac{1}{4(r+1)-1} = \frac{1}{4r-1} - \frac{1}{4r+3} = \frac{4}{(4r-1)(4r+3)}\) | M1; A1 | Clear attempt to use method of differences possibly with one error PI by first A1 |
| (b) \(\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \ldots\) OE or \(f(1) - f(2) + f(2) - f(3) + \ldots\) | M1 | |
| \(\sum_{r=1}^{50}[f(r) - f(r+1)] = f(1) - f(51) = \frac{1}{3} - \frac{1}{203}\) | A1 | |
| \(\sum_{r=1}^{50}\frac{1}{(4r-1)(4r+3)} = \frac{1}{4}\left(\frac{1}{3} - \frac{1}{203}\right)\) | m1 | "their" \(\frac{1}{4}\) × "their" \(\left(\frac{1}{3} - \frac{1}{203}\right)\) |
| \(= \frac{50}{609}\) | A1 | Total: 6 marks |
**(a)** $f(r) - f(r+1) = \frac{1}{4r-1} - \frac{1}{4(r+1)-1} = \frac{1}{4r-1} - \frac{1}{4r+3} = \frac{4}{(4r-1)(4r+3)}$ | M1; A1 | Clear attempt to use method of differences possibly with one error PI by first A1
**(b)** $\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \ldots$ OE or $f(1) - f(2) + f(2) - f(3) + \ldots$ | M1 |
$\sum_{r=1}^{50}[f(r) - f(r+1)] = f(1) - f(51) = \frac{1}{3} - \frac{1}{203}$ | A1 |
$\sum_{r=1}^{50}\frac{1}{(4r-1)(4r+3)} = \frac{1}{4}\left(\frac{1}{3} - \frac{1}{203}\right)$ | m1 | "their" $\frac{1}{4}$ × "their" $\left(\frac{1}{3} - \frac{1}{203}\right)$
$= \frac{50}{609}$ | A1 | **Total: 6 marks**
**Note:** Allow recovery for full marks in part (b) even if errors seen in part (a)
---\begin{enumerate}[label=(\alph*)]
\item Given that $f(r) = \frac{1}{4r-1}$, show that
$$f(r) - f(r+1) = \frac{A}{(4r-1)(4r+3)}$$
where $A$ is an integer. [2 marks]
\item Use the method of differences to find the value of $\sum_{r=1}^{50} \frac{1}{(4r-1)(4r+3)}$, giving your answer as a fraction in its simplest form. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2016 Q1 [6]}}