| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Three linear factors in denominator |
| Difficulty | Standard +0.8 This is a Further Maths question combining partial fractions with method of differences and algebraic manipulation to reach a specific form. Part (a) is routine, but part (b) requires recognizing the telescoping pattern, careful bookkeeping of surviving terms, and non-trivial algebraic manipulation to express the result in the given form with constants a and b. The multi-step nature and requirement to work toward a specified answer format elevates this above standard A-level. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4r+2}{r(r+1)(r+2)} = \frac{1}{r}+\frac{2}{(r+1)}-\frac{3}{(r+2)}\) | M1A1 A1 | M1: Correct partial fractions method (substitution or comparing coefficients) for \(\frac{A}{r},\frac{B}{r+1},\frac{C}{r+2}\); A1: 2 correct fractions; A1: All correct. Correct answer with no working scores full marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n} = \left(\frac{1}{1}+\frac{2}{2}-\frac{3}{3}\right)+\left(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}\right)+\ldots+\left(\frac{1}{n-1}+\frac{2}{n}-\frac{3}{n+1}\right)+\left(\frac{1}{n}+\frac{2}{n+1}-\frac{3}{n+2}\right)\) | M1 | Must have partial fractions of the form \(\frac{A}{r},\frac{B}{r+1},\frac{C}{r+2}\), \(A,B,C\neq 0\). Attempts at least first 2 groups and last 2 groups of terms (may be implied by fractions identified below) |
| \(=\frac{1}{1}+\frac{2}{2}+\frac{1}{2}-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}\) | A1 A1 | A1: \(\frac{1}{1}+\frac{2}{2}+\frac{1}{2}\left(=\frac{5}{2}\right)\) identified as only constant terms; A1: \(-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}\) (or equivalent) as only algebraic terms |
| \(= \frac{5(n^2+3n+2)-2(n+2)-6(n+1)}{2(n+1)(n+2)}\) | M1 | Attempt common denominator from terms of form \(A, \frac{B}{n+1}, \frac{C}{n+2}\) only. Must see \((n+1)(n+2)\) in denominator and unsimplified polynomial of order 2 in numerator |
| \(\dfrac{n(5n+7)}{2(n+1)(n+2)}\) | A1 | Must be in terms of \(n\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{r}+\frac{2}{(r+1)}-\frac{3}{(r+2)}=\left(\frac{1}{r}-\frac{1}{r+2}\right)+2\left(\frac{1}{r+1}-\frac{1}{r+2}\right)\) | — | Rewrite as two telescoping sums |
| \(\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+2}\right)=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\) and \(2\sum_{r=1}^{n}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)=\frac{1}{2}-\frac{1}{n+2}\) | M1 | Re-writes partial fractions correctly and attempts at least 2 groups at start and end for first sum and 1 group at start and end for second sum |
| \(\sum_{r=1}^{n}=\frac{5}{2}-\frac{1}{n+1}-\frac{3}{n+2}\) | A1A1 | A1: \(\frac{1}{1}+\frac{2}{2}+\frac{1}{2}\left(=\frac{5}{2}\right)\) as only constant terms; A1: \(-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}\) or equivalent as only algebraic terms |
| \(= \frac{5(n^2+3n+2)-2(n+2)-6(n+1)}{2(n+1)(n+2)}\) | M1 | Attempt common denominator from terms of form \(A,\frac{B}{n+1},\frac{C}{n+2}\) only. Must see \((n+1)(n+2)\) in denominator and unsimplified polynomial of order 2 in numerator |
| \(\dfrac{n(5n+7)}{2(n+1)(n+2)}\) | A1 | Must be in terms of \(n\) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4r+2}{r(r+1)(r+2)} = \frac{1}{r}+\frac{2}{(r+1)}-\frac{3}{(r+2)}$ | M1A1 A1 | M1: Correct partial fractions method (substitution or comparing coefficients) for $\frac{A}{r},\frac{B}{r+1},\frac{C}{r+2}$; A1: 2 correct fractions; A1: All correct. Correct answer with no working scores full marks. |
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## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} = \left(\frac{1}{1}+\frac{2}{2}-\frac{3}{3}\right)+\left(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}\right)+\ldots+\left(\frac{1}{n-1}+\frac{2}{n}-\frac{3}{n+1}\right)+\left(\frac{1}{n}+\frac{2}{n+1}-\frac{3}{n+2}\right)$ | M1 | **Must have partial fractions of the form** $\frac{A}{r},\frac{B}{r+1},\frac{C}{r+2}$, $A,B,C\neq 0$. Attempts at least first 2 groups and last 2 groups of terms (may be implied by fractions identified below) |
| $=\frac{1}{1}+\frac{2}{2}+\frac{1}{2}-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}$ | A1 A1 | A1: $\frac{1}{1}+\frac{2}{2}+\frac{1}{2}\left(=\frac{5}{2}\right)$ identified as only constant terms; A1: $-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}$ (or equivalent) as only algebraic terms |
| $= \frac{5(n^2+3n+2)-2(n+2)-6(n+1)}{2(n+1)(n+2)}$ | M1 | Attempt common denominator from terms of form $A, \frac{B}{n+1}, \frac{C}{n+2}$ only. Must see $(n+1)(n+2)$ in denominator and unsimplified polynomial of order 2 in numerator |
| $\dfrac{n(5n+7)}{2(n+1)(n+2)}$ | A1 | Must be in terms of $n$ |
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## Question 5(b) Alternative:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{r}+\frac{2}{(r+1)}-\frac{3}{(r+2)}=\left(\frac{1}{r}-\frac{1}{r+2}\right)+2\left(\frac{1}{r+1}-\frac{1}{r+2}\right)$ | — | Rewrite as two telescoping sums |
| $\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+2}\right)=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$ and $2\sum_{r=1}^{n}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)=\frac{1}{2}-\frac{1}{n+2}$ | M1 | Re-writes partial fractions correctly and attempts at least 2 groups at start and end for first sum **and** 1 group at start and end for second sum |
| $\sum_{r=1}^{n}=\frac{5}{2}-\frac{1}{n+1}-\frac{3}{n+2}$ | A1A1 | A1: $\frac{1}{1}+\frac{2}{2}+\frac{1}{2}\left(=\frac{5}{2}\right)$ as only constant terms; A1: $-\frac{3}{n+1}+\frac{2}{n+1}-\frac{3}{n+2}$ or equivalent as only algebraic terms |
| $= \frac{5(n^2+3n+2)-2(n+2)-6(n+1)}{2(n+1)(n+2)}$ | M1 | Attempt common denominator from terms of form $A,\frac{B}{n+1},\frac{C}{n+2}$ only. Must see $(n+1)(n+2)$ in denominator and unsimplified polynomial of order 2 in numerator |
| $\dfrac{n(5n+7)}{2(n+1)(n+2)}$ | A1 | Must be in terms of $n$ |\begin{enumerate}
\item (a) Express $\frac { 4 r + 2 } { r ( r + 1 ) ( r + 2 ) }$ in partial fractions.\\
(b) Hence, using the method of differences, prove that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 4 r + 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( a n + b ) } { 2 ( n + 1 ) ( n + 2 ) }$$
where $a$ and $b$ are constants to be found.
\hfill \mbox{\textit{Edexcel F2 2018 Q5 [8]}}