| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 method of differences question requiring partial fractions decomposition, telescoping sum manipulation, and algebraic simplification to match a given form with unknown constants. While the technique is standard for Further Maths Core Pure, the multi-step algebraic manipulation to reach the specific form and determine a and b requires careful execution, making it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{2}{(r+4)(r+6)} \equiv \frac{A}{r+4} + \frac{B}{r+6} \Rightarrow A = ..., B = ...\) | M1 | Recognises need for partial fractions, applies correct method to find \(A\) and \(B\). Allow a slip when finding constants. |
| \(\frac{2}{(r+4)(r+6)} \equiv \frac{1}{r+4} - \frac{1}{r+6}\) | A1 | Correct partial fractions seen at any stage. Not just values of \(A\) and \(B\) listed. |
| Writing out terms from \(r=1\) and \(r=n\), establishing non-cancelling terms: \(\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n+2}-\frac{1}{n+4}+\frac{1}{n+3}-\frac{1}{n+5}+\frac{1}{n+4}-\frac{1}{n+6}\) | M1 | Starts process of finding terms at start and end to establish non-cancelling terms. Must attempt minimum of \(r=1, r=2, ... r=n-1, r=n\). |
| \(= \frac{1}{5}+\frac{1}{6}-\frac{1}{n+5}-\frac{1}{n+6}\) | A1 | Correct non-cancelling terms which may be listed separately. |
| \(= \frac{1}{5}+\frac{1}{6}-\frac{1}{n+5}-\frac{1}{n+6} = \frac{11(n+5)(n+6)-30(n+6)-30(n+5)}{30(n+5)(n+6)}\) | M1 | Combines fractions of form \(p + \frac{q}{n+5} + \frac{r}{n+6}\) over correct common denominator, obtains quadratic in numerator. |
| \(= \frac{n(11n+61)}{30(n+5)(n+6)}\) | A1 | Correct answer. |
## Question 4:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{2}{(r+4)(r+6)} \equiv \frac{A}{r+4} + \frac{B}{r+6} \Rightarrow A = ..., B = ...$ | M1 | Recognises need for partial fractions, applies correct method to find $A$ and $B$. Allow a slip when finding constants. |
| $\frac{2}{(r+4)(r+6)} \equiv \frac{1}{r+4} - \frac{1}{r+6}$ | A1 | Correct partial fractions seen at any stage. Not just values of $A$ and $B$ listed. |
| Writing out terms from $r=1$ and $r=n$, establishing non-cancelling terms: $\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n+2}-\frac{1}{n+4}+\frac{1}{n+3}-\frac{1}{n+5}+\frac{1}{n+4}-\frac{1}{n+6}$ | M1 | Starts process of finding terms at start and end to establish non-cancelling terms. Must attempt minimum of $r=1, r=2, ... r=n-1, r=n$. |
| $= \frac{1}{5}+\frac{1}{6}-\frac{1}{n+5}-\frac{1}{n+6}$ | A1 | Correct non-cancelling terms which may be listed separately. |
| $= \frac{1}{5}+\frac{1}{6}-\frac{1}{n+5}-\frac{1}{n+6} = \frac{11(n+5)(n+6)-30(n+6)-30(n+5)}{30(n+5)(n+6)}$ | M1 | Combines fractions of form $p + \frac{q}{n+5} + \frac{r}{n+6}$ over correct common denominator, obtains quadratic in numerator. |
| $= \frac{n(11n+61)}{30(n+5)(n+6)}$ | A1 | Correct answer. |
**Note:** Proof by induction will not score the last 4 marks. If starting with $r=0$ maximum score is M1A1M0A0M1A0.
---\begin{enumerate}
\item Use the method of differences to show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 4 ) ( r + 6 ) } = \frac { n ( a n + b ) } { 30 ( n + 5 ) ( n + 6 ) }$$
where $a$ and $b$ are integers to be determined.
\hfill \mbox{\textit{Edexcel CP2 2024 Q4 [6]}}