| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Quadratic factor (difference of squares) |
| Difficulty | Standard +0.3 This is a standard Further Maths partial fractions question with telescoping series. Part (a) is routine factorization of a difference of squares, part (b) is a straightforward telescoping sum proof (a common F2 technique), and part (c) applies the formula with simple arithmetic. While it requires multiple steps, each component follows well-established methods without requiring novel insight. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2(2r-1)} - \frac{1}{2(2r+1)}\) or equivalent; \(A = \frac{1}{2}, B = -\frac{1}{2}\) | B1 | Correct partial fractions or correct values of \(A\) and \(B\). ISW if possible so if correct values found, award when seen even if followed by incorrect partial fractions. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{n} \frac{1}{4r^2-1} = \frac{1}{2}\left(1 - \frac{1}{3} + \cdots + \frac{1}{2n-1} - \frac{1}{2n+1}\right)\) | M1 | Attempt at least first and last terms using their partial fractions. May be implied by e.g. \(\frac{1}{2}\left(1 - \frac{1}{2n+1}\right)\) |
| \(\frac{1}{2}\left(1 - \frac{1}{2n+1}\right)\) or \(\frac{1}{2} - \frac{1}{2(2n+1)}\) or \(\frac{1}{2} - \frac{1}{4n+2}\) | A1 | Correct expression |
| \(\frac{n}{2n+1}\) | A1* | Correct completion with no errors. Allow a different variable in (a) and (b) but final answer must be in terms of \(n\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=9}^{25} \frac{5}{4r^2-1} = (5)(f(25) - f(8))\) | M1 | \(f(25) - f(8)\) where \(f(n) = \frac{n}{2n+1}\) |
| \(= 5\left(\frac{25}{51} - \frac{8}{17}\right) = \frac{5}{51}\) | A1 | cao. Correct answer with no working scores both marks. |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2(2r-1)} - \frac{1}{2(2r+1)}$ or equivalent; $A = \frac{1}{2}, B = -\frac{1}{2}$ | B1 | Correct partial fractions or correct values of $A$ and $B$. ISW if possible so if correct values found, award when seen even if followed by incorrect partial fractions. |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n} \frac{1}{4r^2-1} = \frac{1}{2}\left(1 - \frac{1}{3} + \cdots + \frac{1}{2n-1} - \frac{1}{2n+1}\right)$ | M1 | Attempt at least first and last terms using their partial fractions. May be implied by e.g. $\frac{1}{2}\left(1 - \frac{1}{2n+1}\right)$ |
| $\frac{1}{2}\left(1 - \frac{1}{2n+1}\right)$ or $\frac{1}{2} - \frac{1}{2(2n+1)}$ or $\frac{1}{2} - \frac{1}{4n+2}$ | A1 | Correct expression |
| $\frac{n}{2n+1}$ | A1* | Correct completion with no errors. Allow a different variable in (a) and (b) but final answer must be in terms of $n$. |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=9}^{25} \frac{5}{4r^2-1} = (5)(f(25) - f(8))$ | M1 | $f(25) - f(8)$ where $f(n) = \frac{n}{2n+1}$ |
| $= 5\left(\frac{25}{51} - \frac{8}{17}\right) = \frac{5}{51}$ | A1 | cao. Correct answer with no working scores both marks. |
---\begin{enumerate}
\item (a) Express $\frac { 1 } { 4 r ^ { 2 } - 1 }$ in partial fractions.\\
(b) Hence prove that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }$$
(c) Find the exact value of
$$\sum _ { r = 9 } ^ { 25 } \frac { 5 } { 4 r ^ { 2 } - 1 }$$
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\hfill \mbox{\textit{Edexcel F2 2016 Q1 [6]}}