| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Partial Fractions and Telescoping Series |
| Difficulty | Standard +0.3 This is a standard Further Maths question on series summation using formula booklet and partial fractions with telescoping. Part (a) requires routine expansion and formula application, part (b) is textbook partial fractions leading to a telescoping sum, and part (c) is a direct limit. All techniques are standard with no novel insight required, making it slightly easier than average for Further Maths content. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} r^2 + 2r = \frac{1}{6}n(n+1)(2n+1) + n(n+1)\) | M1 | Uses list of formulae (MF19) |
| \(\frac{1}{6}n(n+1)(2n+7)\) | A1 | At least all like terms collected and fractions simplified |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)\) | M1 A1 | Expresses in partial fractions |
| \(\sum_{r=1}^{n} \frac{2}{r(r+2)} = 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \cdots + \frac{1}{n-1} - \frac{1}{n+1} + \frac{1}{n} - \frac{1}{n+2}\) | M1 | Writes enough terms |
| \(\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)\) | A1 | OE |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{4}\) | B1 FT | FT *their* part (b) of correct form |
| Total: 1 |
## Question 1:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} r^2 + 2r = \frac{1}{6}n(n+1)(2n+1) + n(n+1)$ | M1 | Uses list of formulae (MF19) |
| $\frac{1}{6}n(n+1)(2n+7)$ | A1 | At least all like terms collected and fractions simplified |
| **Total: 2** | | |
---
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$ | M1 A1 | Expresses in partial fractions |
| $\sum_{r=1}^{n} \frac{2}{r(r+2)} = 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \cdots + \frac{1}{n-1} - \frac{1}{n+1} + \frac{1}{n} - \frac{1}{n+2}$ | M1 | Writes enough terms |
| $\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right)$ | A1 | OE |
| **Total: 4** | | |
---
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{4}$ | B1 FT | FT *their* part (b) of correct form |
| **Total: 1** | | |1
\begin{enumerate}[label=(\alph*)]
\item Use the list of formulae (MF19) to find $\sum _ { r = 1 } ^ { n } r ( r + 2 )$ in terms of $n$, simplifying your answer.
\item Express $\frac { 1 } { r ( r + 2 ) }$ in partial fractions and hence find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { \mathrm { r } ( \mathrm { r } + 2 ) }$ in terms of $n$.
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 2 ) }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q1 [7]}}