| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question combining routine partial fractions with telescoping series summation. Part (a) is standard A-level technique, and part (b) requires recognizing the telescoping pattern and simplifying—methodical but not conceptually demanding for FP2 students. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2}{(r+2)(r+4)} = \frac{1}{r+2} - \frac{1}{r+4}\) | B1 (1) | Correct partial fractions. Can be seen in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{n} \frac{2}{(r+2)(r+4)} = \sum_{r=1}^{n}\left(\frac{1}{r+2} - \frac{1}{r+4}\right)\), showing first 2 and last 2 terms | M1 | Attempts at least first 2 and last 2 terms. Must start at 1 and end at \(n\) |
| \(= \frac{1}{3} + \frac{1}{4} - \frac{1}{n+3} - \frac{1}{n+4}\) | M1A1 | M1: Identifies four fractions that do not cancel. If all terms positive this mark is lost. A1: Correct four fractions |
| \(= \frac{7}{12} - \frac{1}{n+3} - \frac{1}{n+4}\) | ||
| \(= \frac{7(n+3)(n+4) - 12(n+4) - 12(n+3)}{12(n+3)(n+4)}\) | M1 | Attempt to combine at least 3 fractions, 2 of which have a function of \(n\) in denominator, and expands numerator. Product of 2 linear factors must be expanded |
| \(= \frac{n(7n+25)}{12(n+3)(n+4)}\) | A1 (5) | Must be factorised. If worked with \(r\) instead of \(n\) throughout, deduct last mark only |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2}{(r+2)(r+4)} = \frac{1}{r+2} - \frac{1}{r+4}$ | B1 (1) | Correct partial fractions. Can be seen in (b) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n} \frac{2}{(r+2)(r+4)} = \sum_{r=1}^{n}\left(\frac{1}{r+2} - \frac{1}{r+4}\right)$, showing first 2 and last 2 terms | M1 | Attempts at least first 2 and last 2 terms. Must start at 1 and end at $n$ |
| $= \frac{1}{3} + \frac{1}{4} - \frac{1}{n+3} - \frac{1}{n+4}$ | M1A1 | M1: Identifies four fractions that do not cancel. If all terms positive this mark is lost. A1: Correct four fractions |
| $= \frac{7}{12} - \frac{1}{n+3} - \frac{1}{n+4}$ | | |
| $= \frac{7(n+3)(n+4) - 12(n+4) - 12(n+3)}{12(n+3)(n+4)}$ | M1 | Attempt to combine at least 3 fractions, 2 of which have a function of $n$ in denominator, and expands numerator. Product of 2 linear factors must be expanded |
| $= \frac{n(7n+25)}{12(n+3)(n+4)}$ | A1 (5) | Must be factorised. If worked with $r$ instead of $n$ throughout, deduct last mark only |
---\begin{enumerate}
\item (a) Express $\frac { 2 } { ( r + 2 ) ( r + 4 ) }$ in partial fractions.\\
(b) Hence show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 2 ) ( r + 4 ) } = \frac { n ( 7 n + 25 ) } { 12 ( n + 3 ) ( n + 4 ) }$$
\hfill \mbox{\textit{Edexcel FP2 2014 Q1 [6]}}