| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Partial fractions then method of differences |
| Difficulty | Standard +0.3 This is a straightforward application of two standard A-level techniques: partial fractions decomposition (routine for linear factors) followed by a telescoping series. While it requires careful bookkeeping in part (b), both methods are well-practiced in FP2 with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{r(r+2)} = \frac{A}{r} + \frac{B}{r+2}\), so \(1 = A(r+2) + Br\) | M1 | Correct method |
| \(A = \frac{1}{2},\ B = -\frac{1}{2}\) | A1 | Both correct |
| \(\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)\) | A1 | Correct form |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{48} \frac{1}{r(r+2)} = \frac{1}{2}\sum_{r=1}^{48}\left(\frac{1}{r} - \frac{1}{r+2}\right)\) | M1 | Using partial fractions |
| \(= \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \cdots\right]\) | M1 | Writing out terms to show cancellation |
| Surviving terms: \(\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{49} - \frac{1}{50}\right)\) | A1 | Correct surviving terms |
| \(= \frac{1}{2}\left(\frac{3}{2} - \frac{99}{2450}\right) = \frac{1}{2} \cdot \frac{3576}{2450}\) | M1 | Correct simplification |
| \(= \frac{1788}{2450} = \frac{894}{1225}\) | A1 | Correct rational answer |
# Question 2:
## Part (a):
| $\frac{1}{r(r+2)} = \frac{A}{r} + \frac{B}{r+2}$, so $1 = A(r+2) + Br$ | M1 | Correct method |
|---|---|---|
| $A = \frac{1}{2},\ B = -\frac{1}{2}$ | A1 | Both correct |
| $\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$ | A1 | Correct form |
## Part (b):
| $\sum_{r=1}^{48} \frac{1}{r(r+2)} = \frac{1}{2}\sum_{r=1}^{48}\left(\frac{1}{r} - \frac{1}{r+2}\right)$ | M1 | Using partial fractions |
|---|---|---|
| $= \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \cdots\right]$ | M1 | Writing out terms to show cancellation |
| Surviving terms: $\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{49} - \frac{1}{50}\right)$ | A1 | Correct surviving terms |
| $= \frac{1}{2}\left(\frac{3}{2} - \frac{99}{2450}\right) = \frac{1}{2} \cdot \frac{3576}{2450}$ | M1 | Correct simplification |
| $= \frac{1788}{2450} = \frac{894}{1225}$ | A1 | Correct rational answer |
---2 (a) Express $\frac { 1 } { r ( r + 2 ) }$ in partial fractions.\\
(b) Use the method of differences to find
$$\sum _ { r = 1 } ^ { 48 } \frac { 1 } { r ( r + 2 ) }$$
giving your answer as a rational number.
\hfill \mbox{\textit{AQA FP2 2010 Q2 [8]}}