| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Standard +0.3 This is a standard Further Maths partial fractions question with method of differences. Part (a) is routine decomposition, part (b) follows a well-practiced telescoping technique with algebraic manipulation to match the given form, and part (c) is direct substitution. While it requires multiple steps and careful algebra, it follows predictable patterns that Further Maths students drill extensively, making it slightly easier than average overall. |
| Spec | 1.02y Partial fractions: decompose rational functions4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(n+3)(n+5)} \equiv \frac{A}{(n+3)}+\frac{B}{(n+5)} \Rightarrow A=\ldots, B=\ldots\) | M1 | Correct partial fraction attempt to obtain values for \(A\) and \(B\) |
| \(\frac{1}{2(n+3)}-\frac{1}{2(n+5)}\) | A1 | Correct expression. May have \(\frac{1}{2}\) in each numerator or factored out |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r+3}-\frac{1}{r+5}\right) = \frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\ldots+\frac{1}{n+2}-\frac{1}{n+4}+\frac{1}{n+3}-\frac{1}{n+5}\right)\) | M1 | Uses method of differences to identify non-cancelling terms. Expect at least first two pairs and last pair listed |
| \(=\frac{1}{2}\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{n+4}-\frac{1}{n+5}\right)\) | A1 | Identifies correct non-cancelling terms |
| \(=\frac{1}{2}\left(\frac{9(n+4)(n+5)-20(n+5)-20(n+4)}{20(n+4)(n+5)}\right)\) | dM1 | Puts over common denominator \(k(n+4)(n+5)\), \(k\neq 1\), progresses towards required form |
| \(=\frac{n(9n+41)}{40(n+4)(n+5)}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{9\times11}+\frac{1}{10\times12}+\ldots+\frac{1}{24\times26}=\frac{21(9\times21+41)}{40\times25\times26}-\frac{5(9\times5+41)}{40\times9\times10}\) | M1 | Attempts to use formula from (b) with \(\sum_{r=1}^{21}-\sum_{r=1}^{k}\) where \(k=5\) or \(6\). Substitution must be seen at least once |
| \(=\frac{194}{2925}\) | A1 | Correct answer achieved (M must have been scored). Must be simplified |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(n+3)(n+5)} \equiv \frac{A}{(n+3)}+\frac{B}{(n+5)} \Rightarrow A=\ldots, B=\ldots$ | M1 | Correct partial fraction attempt to obtain values for $A$ and $B$ |
| $\frac{1}{2(n+3)}-\frac{1}{2(n+5)}$ | A1 | Correct expression. May have $\frac{1}{2}$ in each numerator or factored out |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r+3}-\frac{1}{r+5}\right) = \frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\ldots+\frac{1}{n+2}-\frac{1}{n+4}+\frac{1}{n+3}-\frac{1}{n+5}\right)$ | M1 | Uses method of differences to identify non-cancelling terms. Expect at least first two pairs and last pair listed |
| $=\frac{1}{2}\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{n+4}-\frac{1}{n+5}\right)$ | A1 | Identifies correct non-cancelling terms |
| $=\frac{1}{2}\left(\frac{9(n+4)(n+5)-20(n+5)-20(n+4)}{20(n+4)(n+5)}\right)$ | dM1 | Puts over common denominator $k(n+4)(n+5)$, $k\neq 1$, progresses towards required form |
| $=\frac{n(9n+41)}{40(n+4)(n+5)}$ | A1 | Correct answer |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{9\times11}+\frac{1}{10\times12}+\ldots+\frac{1}{24\times26}=\frac{21(9\times21+41)}{40\times25\times26}-\frac{5(9\times5+41)}{40\times9\times10}$ | M1 | Attempts to use formula from (b) with $\sum_{r=1}^{21}-\sum_{r=1}^{k}$ where $k=5$ or $6$. Substitution must be seen at least once |
| $=\frac{194}{2925}$ | A1 | Correct answer achieved (M must have been scored). Must be simplified |
---\begin{enumerate}
\item (a) Express
\end{enumerate}
$$\frac { 1 } { ( n + 3 ) ( n + 5 ) }$$
in partial fractions.\\
(b) Hence, using the method of differences, show that for all positive integer values of $n$,
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 3 ) ( r + 5 ) } = \frac { n ( p n + q ) } { 40 ( n + 4 ) ( n + 5 ) }$$
where $p$ and $q$ are integers to be determined.\\
(c) Use the answer to part (b) to determine, as a simplified fraction, the value of
$$\frac { 1 } { 9 \times 11 } + \frac { 1 } { 10 \times 12 } + \ldots + \frac { 1 } { 24 \times 26 }$$
\hfill \mbox{\textit{Edexcel F2 2024 Q3 [8]}}