| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated squared factors in denominator |
| Difficulty | Standard +0.8 This is a Further Maths F2 question requiring partial fractions with repeated factors (harder than standard A-level), followed by telescoping series summation. The partial fractions setup is non-standard as it omits the linear terms, requiring students to recognize this works due to the specific numerator. The summation requires careful algebraic manipulation to reach the given form. More challenging than typical C4 partial fractions but standard for F2. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2n+1 = A(n+1)^2 + Bn^2 \Rightarrow 2n+1 = An^2 + 2An + 1 + Bn^2\) | ||
| \(A = 1\), \(B = -1\) or \(\frac{1}{n^2} - \frac{1}{(n+1)^2}\) | B1 (1) | Both values correct with or without working. Ignore incorrect working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=5}^{n}\left(\frac{1}{r^2} - \frac{1}{(r+1)^2}\right)\), showing telescoping terms | M1 | Show sufficient terms to demonstrate cancelling. At least one cancelling term seen. Must start at \(r=5\). |
| \(\sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2} = \frac{1}{5^2} - \frac{1}{(n+1)^2}\) | A1 | Extract the two correct remaining terms. |
| \(= \frac{n^2+2n+1-25}{25(n+1)^2} = \frac{n^2+2n-24}{25(n+1)^2}\) | M1A1 (4) | Write with common denominator, numerator correct. Correct final answer. |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2n+1 = A(n+1)^2 + Bn^2 \Rightarrow 2n+1 = An^2 + 2An + 1 + Bn^2$ | | |
| $A = 1$, $B = -1$ or $\frac{1}{n^2} - \frac{1}{(n+1)^2}$ | B1 (1) | Both values correct with or without working. Ignore incorrect working. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2} = \sum_{r=5}^{n}\left(\frac{1}{r^2} - \frac{1}{(r+1)^2}\right)$, showing telescoping terms | M1 | Show sufficient terms to demonstrate cancelling. At least one cancelling term seen. Must start at $r=5$. |
| $\sum_{r=5}^{n} \frac{2r+1}{r^2(r+1)^2} = \frac{1}{5^2} - \frac{1}{(n+1)^2}$ | A1 | Extract the two correct remaining terms. |
| $= \frac{n^2+2n+1-25}{25(n+1)^2} = \frac{n^2+2n-24}{25(n+1)^2}$ | M1A1 (4) | Write with common denominator, numerator correct. Correct final answer. |
---\begin{enumerate}
\item Given that
\end{enumerate}
$$\frac { 2 n + 1 } { n ^ { 2 } ( n + 1 ) ^ { 2 } } \equiv \frac { A } { n ^ { 2 } } + \frac { B } { ( n + 1 ) ^ { 2 } }$$
(a) determine the value of $A$ and the value of $B$\\
(b) Hence show that, for $n \geqslant 5$
$$\sum _ { r = 5 } ^ { n } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = \frac { n ^ { 2 } + a n + b } { c ( n + 1 ) ^ { 2 } }$$
where $a$, $b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2022 Q1 [5]}}