| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a standard method of differences question with the partial fraction decomposition already provided. Part (a) requires routine algebraic verification, and part (b) involves telescoping the series—a well-practiced technique in Further Maths. While it requires careful bookkeeping of terms, it follows a familiar pattern with no novel insight needed. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2(r+1)(r+2)} - \frac{1}{2(r+2)(r+3)} = \frac{r+3-(r+1)}{2(r+1)(r+2)(r+3)}\) | M1 | Attempt common denominator |
| \(= \frac{2}{2(r+1)(r+2)(r+3)} = \frac{1}{(r+1)(r+2)(r+3)}\) | A1 | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(r+1)(r+2)(r+3)} = \frac{1}{r+2}\left(\frac{1}{(r+1)(r+3)}\right) = \frac{1}{r+2}\left(\frac{1}{2(r+1)}-\frac{1}{2(r+3)}\right)\) | M1 | Factor of \(\frac{1}{r+2}\) and attempt partial fractions |
| \(\frac{1}{2(r+1)(r+2)} - \frac{1}{2(r+2)(r+3)}\) | A1 | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)(r+3)} = \frac{1}{12} - \frac{1}{24} + \ldots + \frac{1}{2(n+1)(n+2)} - \frac{1}{2(n+2)(n+3)}\) | M1 | Attempt at least the first pair and the last pair of terms as shown. Must start at 1 and end at \(n\) |
| \(= \frac{1}{12} - \frac{1}{2(n+2)(n+3)}\) | M1 | Identifies that the first and last terms do not cancel |
| \(= \frac{n^2+5n+6-6}{12(n+2)(n+3)}\) | A1 | Correctly combined fractions |
| \(= \frac{n^2+5n}{12(n+2)(n+3)}\) or \(\frac{n(n+5)}{12(n+2)(n+3)}\) | A1 | Allow either form; isw attempts to multiply out the denominator |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2(r+1)(r+2)} - \frac{1}{2(r+2)(r+3)} = \frac{r+3-(r+1)}{2(r+1)(r+2)(r+3)}$ | M1 | Attempt common denominator |
| $= \frac{2}{2(r+1)(r+2)(r+3)} = \frac{1}{(r+1)(r+2)(r+3)}$ | A1 | Correct proof |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(r+1)(r+2)(r+3)} = \frac{1}{r+2}\left(\frac{1}{(r+1)(r+3)}\right) = \frac{1}{r+2}\left(\frac{1}{2(r+1)}-\frac{1}{2(r+3)}\right)$ | M1 | Factor of $\frac{1}{r+2}$ and attempt partial fractions |
| $\frac{1}{2(r+1)(r+2)} - \frac{1}{2(r+2)(r+3)}$ | A1 | Correct proof |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)(r+3)} = \frac{1}{12} - \frac{1}{24} + \ldots + \frac{1}{2(n+1)(n+2)} - \frac{1}{2(n+2)(n+3)}$ | M1 | Attempt at least the first pair and the last pair of terms as shown. Must start at 1 and end at $n$ |
| $= \frac{1}{12} - \frac{1}{2(n+2)(n+3)}$ | M1 | Identifies that the first and last terms do not cancel |
| $= \frac{n^2+5n+6-6}{12(n+2)(n+3)}$ | A1 | Correctly combined fractions |
| $= \frac{n^2+5n}{12(n+2)(n+3)}$ or $\frac{n(n+5)}{12(n+2)(n+3)}$ | A1 | Allow either form; isw attempts to multiply out the denominator |
---\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$\frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } \equiv \frac { 1 } { 2 ( r + 1 ) ( r + 2 ) } - \frac { 1 } { 2 ( r + 2 ) ( r + 3 ) }$$
(b) Hence, or otherwise, find
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }$$
giving your answer as a single fraction in its simplest form.\\
\hfill \mbox{\textit{Edexcel F2 2014 Q1 [6]}}