| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Three linear factors in denominator |
| Difficulty | Standard +0.3 This is a standard Further Maths partial fractions question with telescoping series. Part (a) is routine decomposition with three linear factors. Part (b) requires recognizing the telescoping pattern, which is a well-practiced technique in F2. Part (c) is straightforward substitution. While it's a multi-part question requiring several steps, all techniques are standard textbook exercises for Further Maths students with no novel insight required. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| VIXV SIHII NI JIIIM ION OC | VIAN SIHI NI JYHM ION OO | VAYV SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{3r+1}{r(r-1)(r+1)} = \frac{A}{r} + \frac{B}{r-1} + \frac{C}{r+1}\) | M1 | Correct method for obtaining the partial fractions |
| \(\frac{3r+1}{r(r-1)(r+1)} = -\frac{1}{r} + \frac{2}{r-1} - \frac{1}{r+1}\) | A1 (2) | Correct PFs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Telescoping table of terms shown | M1 | Show sufficient terms at both ends (e.g. 3 at start and 2 at end) to demonstrate cancelling. Must use PFs of correct form starting at \(r=2\) unless extra terms ignored. Can be split into \(\sum\left(\frac{1}{r-1}-\frac{1}{r}\right)+\sum\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\) |
| \(= 2 - \frac{1}{2} + \frac{2}{2} - \frac{1}{n} - \frac{1}{n} - \frac{1}{n+1}\) | dM1A1 | Extract non-cancelled terms (min 4 correct terms but 5/2 counts as 3 correct). Depends on first M of (b). A1: Correct terms extracted |
| \(\frac{5}{2} - \frac{2}{n} - \frac{1}{n+1} = \frac{5n(n+1)-4(n+1)-2n}{2n(n+1)} = \frac{5n^2-n-4}{2n(n+1)}\) | M1, A1 cso (5) | M1: Write terms using common denominator; numerator need not be simplified. Must start with min of 3 terms inc terms with denominators \(n\) and \((n+1)\). A1: Correct answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{2}^{20} - \sum_{2}^{14}\) | M1 | Form and use the difference of the 2 summations shown using result from (b) or earlier form seen in (b) |
| \(= \frac{5\times20^2-20-4}{2\times20\times21} - \frac{5\times14^2-14-4}{2\times14\times15}\) | M1 | |
| \(= \frac{13}{210}\) | A1 (2) | Correct exact answer, as shown or equivalent |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3r+1}{r(r-1)(r+1)} = \frac{A}{r} + \frac{B}{r-1} + \frac{C}{r+1}$ | M1 | Correct method for obtaining the partial fractions |
| $\frac{3r+1}{r(r-1)(r+1)} = -\frac{1}{r} + \frac{2}{r-1} - \frac{1}{r+1}$ | A1 (2) | Correct PFs |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Telescoping table of terms shown | M1 | Show sufficient terms at both ends (e.g. 3 at start and 2 at end) to demonstrate cancelling. Must use PFs of correct form starting at $r=2$ unless extra terms ignored. Can be split into $\sum\left(\frac{1}{r-1}-\frac{1}{r}\right)+\sum\left(\frac{1}{r-1}-\frac{1}{r+1}\right)$ |
| $= 2 - \frac{1}{2} + \frac{2}{2} - \frac{1}{n} - \frac{1}{n} - \frac{1}{n+1}$ | dM1A1 | Extract non-cancelled terms (min 4 correct terms but 5/2 counts as 3 correct). Depends on first M of (b). A1: Correct terms extracted |
| $\frac{5}{2} - \frac{2}{n} - \frac{1}{n+1} = \frac{5n(n+1)-4(n+1)-2n}{2n(n+1)} = \frac{5n^2-n-4}{2n(n+1)}$ | M1, A1 cso (5) | M1: Write terms using common denominator; numerator need not be simplified. Must start with min of 3 terms inc terms with denominators $n$ and $(n+1)$. A1: Correct answer from correct working |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{2}^{20} - \sum_{2}^{14}$ | M1 | Form and use the difference of the 2 summations shown using result from (b) or earlier form seen in (b) |
| $= \frac{5\times20^2-20-4}{2\times20\times21} - \frac{5\times14^2-14-4}{2\times14\times15}$ | M1 | |
| $= \frac{13}{210}$ | A1 (2) | Correct exact answer, as shown or equivalent |
**Total: [9]**
---2. (a) Write $\frac { 3 r + 1 } { r ( r - 1 ) ( r + 1 ) }$ in partial fractions.\\
(b) Hence find
$$\sum _ { r = 2 } ^ { n } \frac { 3 r + 1 } { r ( r - 1 ) ( r + 1 ) } \quad n \geqslant 2$$
giving your answer in the form
$$\frac { a n ^ { 2 } + b n + c } { 2 n ( n + 1 ) }$$
where $a$, $b$ and $c$ are integers to be determined.\\
(c) Hence determine the exact value of
$$\sum _ { r = 15 } ^ { 20 } \frac { 3 r + 1 } { r ( r - 1 ) ( r + 1 ) }$$
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VIXV SIHII NI JIIIM ION OC & VIAN SIHI NI JYHM ION OO & VAYV SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel F2 2020 Q2 [9]}}