| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.8 This is a standard Further Maths method of differences question requiring partial fraction decomposition and telescoping series summation. While it involves multiple steps (finding constants, applying the method, handling non-standard limits), these are well-practiced techniques for FP2 students with no novel insight required. The arithmetic with r=10 to r=98 adds minor complexity but remains routine. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) \(1 = A(r+2) + Br\) | M1 | |
| \(2A = 1, A = \frac{1}{2}\) | A1 | |
| \(A + B = 0, B = -\frac{1}{2}\) | A1 | 3 marks |
| if (a) is incorrect but \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\) used, allow full marks for (b) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 98\): \(\frac{1}{2}\left(\frac{1}{98.99} - \frac{1}{99.100}\right)\) | M1A1 | 3 relevant rows seen |
| \(S = \frac{1}{2}\left(\frac{1}{10.11} - \frac{1}{99.100}\right)\) | m1 | if split into \(\frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}\), follow mark scheme, which case \(\frac{1}{2.10} - \frac{1}{2.11} + \frac{1}{2.100} - \frac{1}{2.99}\) scores m1 |
| \(= \frac{89}{19800}\) | A1 | 4 marks (7 marks for question) |
**2(a)** $1 = A(r+2) + Br$ | M1
$2A = 1, A = \frac{1}{2}$ | A1
$A + B = 0, B = -\frac{1}{2}$ | A1 | 3 marks
| | | if (a) is incorrect but $A = \frac{1}{2}$ and $B = -\frac{1}{2}$ used, allow full marks for (b)
**2(b)** $r = 10$: $\frac{1}{2}\left(\frac{1}{10.11} - \frac{1}{11.12}\right)$
$r = 11$: $\frac{1}{2}\left(\frac{1}{11.12} - \frac{1}{12.13}\right)$
...
$r = 98$: $\frac{1}{2}\left(\frac{1}{98.99} - \frac{1}{99.100}\right)$ | M1A1 | 3 relevant rows seen
$S = \frac{1}{2}\left(\frac{1}{10.11} - \frac{1}{99.100}\right)$ | m1 | if split into $\frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}$, follow mark scheme, which case $\frac{1}{2.10} - \frac{1}{2.11} + \frac{1}{2.100} - \frac{1}{2.99}$ scores m1
$= \frac{89}{19800}$ | A1 | 4 marks (7 marks for question)2
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\frac { 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { A } { r ( r + 1 ) } + \frac { B } { ( r + 1 ) ( r + 2 ) }$$
show that $A = \frac { 1 } { 2 }$ and find the value of $B$.
\item Use the method of differences to find
$$\sum _ { r = 10 } ^ { 98 } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$$
giving your answer as a rational number.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q2 [7]}}