| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| 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 straightforward method of differences question where part (a) is routine algebraic verification requiring only finding a common denominator, and part (b) involves recognizing that 2r+1 fits the given identity, then applying telescoping series—a standard FP2 technique with minimal problem-solving required beyond pattern recognition. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2}\) | M1 | |
| \(= \frac{2r+1}{r^2(r+1)^2}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{3}{1^2 \times 2^2} = \frac{1}{1^2} - \frac{1}{2^2}\), \(\frac{5}{2^2 \times 3^2} = \frac{1}{2^2} - \frac{1}{3^2}\), \(\frac{7}{3^2 \times 4^2} = \frac{1}{3^2} - \frac{1}{4^2}\) | M1A1 | A1 for at least 3 lines |
| \(\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}\) | ||
| Clear cancellation | M1 | |
| \(1 - \frac{1}{(n+1)^2}\) | A1F |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2}$ | M1 | |
| $= \frac{2r+1}{r^2(r+1)^2}$ | A1 | AG |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3}{1^2 \times 2^2} = \frac{1}{1^2} - \frac{1}{2^2}$, $\frac{5}{2^2 \times 3^2} = \frac{1}{2^2} - \frac{1}{3^2}$, $\frac{7}{3^2 \times 4^2} = \frac{1}{3^2} - \frac{1}{4^2}$ | M1A1 | A1 for at least 3 lines |
| $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ | | |
| Clear cancellation | M1 | |
| $1 - \frac{1}{(n+1)^2}$ | A1F | |
---1
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 1 ) ^ { 2 } } = \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }$$
\item Hence find the sum of the first $n$ terms of the series
$$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q1 [6]}}