| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Standard +0.8 This is a Further Maths question combining standard partial fractions with telescoping series summation, requiring algebraic manipulation to reach a specific form. While partial fractions with two linear factors is routine, the summation and simplification to match the given expression requires careful algebraic work and insight into telescoping series, placing it moderately above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2(r+6)} - \frac{1}{2(r+8)}\) | B1 | Correct partial fractions, any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \left(2 \times \frac{1}{2}\right)\left(\frac{1}{7} - \frac{1}{9} + \frac{1}{8} - \frac{1}{10} + \frac{1}{9} - \frac{1}{11} \cdots + \frac{1}{n+5} - \frac{1}{n+7} + \frac{1}{n+6} - \frac{1}{n+8}\right)\) | M1 | Expands at least 3 terms at start and 2 at end. Partial fractions from (a) can be used without multiplying by 2. Fractions may be \(\frac{1}{2}\times\frac{1}{7} - \frac{1}{2}\times\frac{1}{9}\) etc |
| \(= \frac{1}{7} + \frac{1}{8} - \frac{1}{n+7} - \frac{1}{n+8}\) | A1 | Identifies the terms that do not cancel |
| \(= \frac{15(n+7)(n+8) - 56(2n+15)}{56(n+7)(n+8)}\) | M1 | Attempt common denominator. Must have multiplied fractions from (a) by 2 now |
| \(= \frac{n(15n+113)}{56(n+7)(n+8)}\) | A1cso |
## Question 2:
$$\frac{1}{(r+6)(r+8)}$$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2(r+6)} - \frac{1}{2(r+8)}$ | B1 | Correct partial fractions, any equivalent form |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \left(2 \times \frac{1}{2}\right)\left(\frac{1}{7} - \frac{1}{9} + \frac{1}{8} - \frac{1}{10} + \frac{1}{9} - \frac{1}{11} \cdots + \frac{1}{n+5} - \frac{1}{n+7} + \frac{1}{n+6} - \frac{1}{n+8}\right)$ | M1 | Expands at least 3 terms at start and 2 at end. Partial fractions from (a) can be used without multiplying by 2. Fractions may be $\frac{1}{2}\times\frac{1}{7} - \frac{1}{2}\times\frac{1}{9}$ etc |
| $= \frac{1}{7} + \frac{1}{8} - \frac{1}{n+7} - \frac{1}{n+8}$ | A1 | Identifies the terms that do not cancel |
| $= \frac{15(n+7)(n+8) - 56(2n+15)}{56(n+7)(n+8)}$ | M1 | Attempt common denominator. Must have multiplied fractions from (a) by 2 now |
| $= \frac{n(15n+113)}{56(n+7)(n+8)}$ | A1cso | |
---\begin{enumerate}
\item (a) Express $\frac { 1 } { ( r + 6 ) ( r + 8 ) }$ in partial fractions.\\
(b) Hence show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 6 ) ( r + 8 ) } = \frac { n ( a n + b ) } { 56 ( n + 7 ) ( n + 8 ) }$$
where $a$ and $b$ are integers to be found.\\
\hfill \mbox{\textit{Edexcel F2 2015 Q2 [5]}}