| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| 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 partial fractions with telescoping series summation. Part (a) is routine, but part (b) requires recognizing the telescoping pattern, performing the summation correctly, and algebraic manipulation to match the given form with specific coefficients. The multi-step nature and need for careful algebraic work elevates this above average difficulty. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| VIIN SIHI NI JIIIM IONOO | VIIV SIHI NI III HM ION OO | VI4V SIHI NI JIIIM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(r+6)(r+8)}\) | Given expression | |
| \(\frac{1}{2(r+6)} - \frac{1}{2(r+8)}\) | B1 | Correct partial fractions, any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \left(2 \times \frac{1}{2}\right)\left(\frac{1}{7} - \frac{1}{9} + \frac{1}{8} - \frac{1}{10} + \frac{1}{9} - \frac{1}{11}... + \frac{1}{n+5} - \frac{1}{n+7} + \frac{1}{n+6} - \frac{1}{n+8}\right)\) | M1 A1 | Expands at least 3 terms at start and 2 at end (may be implied). 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 the fractions from (a) by 2 now |
| \(= \frac{n(15n+113)}{56(n+7)(n+8)}\) | A1 cso |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(r+6)(r+8)}$ | | Given expression |
| $\frac{1}{2(r+6)} - \frac{1}{2(r+8)}$ | B1 | Correct partial fractions, any equivalent form |
**Total: (1)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \left(2 \times \frac{1}{2}\right)\left(\frac{1}{7} - \frac{1}{9} + \frac{1}{8} - \frac{1}{10} + \frac{1}{9} - \frac{1}{11}... + \frac{1}{n+5} - \frac{1}{n+7} + \frac{1}{n+6} - \frac{1}{n+8}\right)$ | M1 A1 | Expands at least 3 terms at start and 2 at end (may be implied). 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 the fractions from (a) by 2 now |
| $= \frac{n(15n+113)}{56(n+7)(n+8)}$ | A1 cso | |
**Total: (4)**
---\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.\\
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VIIN SIHI NI JIIIM IONOO & VIIV SIHI NI III HM ION OO & VI4V SIHI NI JIIIM IONOO \\
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\hfill \mbox{\textit{Edexcel F2 2018 Q2 [5]}}