| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Challenging +1.2 This is a structured method-of-differences question with explicit guidance. Part (i) is routine algebraic manipulation (1 mark). Part (ii) requires recognizing how to apply the telescoping sum and simplifying, but the method is named and the structure is given. While it involves more steps than a basic question, the technique is standard for Further Maths students and the question provides significant scaffolding. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(r-1) - f(r) = \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} = \frac{8r}{(4r^2-1)}\) (denom'. may be factorised) | B1 | |
| [1] | ||
| (ii) \(\sum_{r=1}^{n} \frac{r}{(4r^2-1)^2} = \frac{1}{8}\sum_{r=1}^{n}\{f(r-1)-f(r)\}\) | M1 | Use of this result … |
| \(= \frac{1}{8}\sum_{r=1}^{n}\{(f(0)-f(1))+(f(1)-f(2))+\cdots+(f(n-1)-f(n))\}\) | M1 | & cancelling |
| \(= \frac{1}{8}\{f(0)-f(n)\} = \frac{1}{8}\left(1-\frac{1}{(2n+1)^2}\right)\) | A1 | |
| \(= \frac{1}{8}\left(\frac{(4n^2+4n+1)-1}{(2n+1)^2}\right)\) | A1 | Common denom'. with squaring attempted in num'. |
| \(= \frac{n(n+1)}{2(2n+1)^2}\) GIVEN ANSWER from correct working | A1 | |
| [4] |
**(i)** $f(r-1) - f(r) = \frac{1}{(2r-1)^2} - \frac{1}{(2r+1)^2} = \frac{8r}{(4r^2-1)}$ (denom'. may be factorised) | B1 |
| [1] |
**(ii)** $\sum_{r=1}^{n} \frac{r}{(4r^2-1)^2} = \frac{1}{8}\sum_{r=1}^{n}\{f(r-1)-f(r)\}$ | M1 | Use of this result …
$= \frac{1}{8}\sum_{r=1}^{n}\{(f(0)-f(1))+(f(1)-f(2))+\cdots+(f(n-1)-f(n))\}$ | M1 | & cancelling
$= \frac{1}{8}\{f(0)-f(n)\} = \frac{1}{8}\left(1-\frac{1}{(2n+1)^2}\right)$ | A1 |
$= \frac{1}{8}\left(\frac{(4n^2+4n+1)-1}{(2n+1)^2}\right)$ | A1 | Common denom'. with squaring attempted in num'.
$= \frac{n(n+1)}{2(2n+1)^2}$ **GIVEN ANSWER from correct working** | A1 |
| [4] |
---\begin{enumerate}[label=(\roman*)]
\item Express $\text{f}(r - 1) - \text{f}(r)$ as a single algebraic fraction, where $\text{f}(r) = \frac{1}{(2r + 1)^2}$. [1]
\item Hence, using the method of differences, show that
$$\sum_{r=1}^{n} \frac{r}{(4r^2 - 1)^2} = \frac{n(n + 1)}{2(2n + 1)^2}$$
for all positive integers $n$. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q3 [5]}}