| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Challenging +1.2 This is a standard Further Maths method of differences question with straightforward partial fractions. Part (a) requires routine telescoping, (b) is immediate from taking the limit, and (c) applies the formula from (a) to find the difference between two sums. While it involves multiple parts and algebraic manipulation, the techniques are well-practiced and the question provides clear scaffolding with no novel insights required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\left(\frac{1}{k(r-1)+1} - \frac{1}{kr+1}\right)\) | M1 A1 | Finds partial fractions. |
| \(\sum_{r=1}^{n}\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\left(1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \cdots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1}\right)\) | M1 | Writes at least three complete terms, including last. |
| \(\frac{1}{k}\left(1-\frac{1}{kn+1}\right)\) | A1 | OE e.g. \(\frac{n}{kn+1}\) |
| Answer | Marks |
|---|---|
| \(\frac{1}{k}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=n}^{n^2}\frac{1}{(kr+1)(kr-k+1)} = \sum_{r=1}^{n^2}\frac{1}{(kr+1)(kr-k+1)} - \sum_{r=1}^{n-1}\frac{1}{(kr+1)(kr-k+1)}\) | M1 | Or applies the method of differences again. |
| \(\frac{1}{k}\left(1-\frac{1}{kn^2+1}-\left(1-\frac{1}{k(n-1)+1}\right)\right) = \frac{1}{k}\left(\frac{1}{k(n-1)+1}-\frac{1}{kn^2+1}\right)\) | A1 | OE e.g. \(\frac{n^2}{kn^2+1} - \frac{(n-1)}{k(n-1)+1}\) |
## Question 3:
**Part 3(a):**
$\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\left(\frac{1}{k(r-1)+1} - \frac{1}{kr+1}\right)$ | M1 A1 | Finds partial fractions.
$\sum_{r=1}^{n}\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\left(1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \cdots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1}\right)$ | M1 | Writes at least three complete terms, including last.
$\frac{1}{k}\left(1-\frac{1}{kn+1}\right)$ | A1 | OE e.g. $\frac{n}{kn+1}$
**Part 3(b):**
$\frac{1}{k}$ | B1 |
**Part 3(c):**
$\sum_{r=n}^{n^2}\frac{1}{(kr+1)(kr-k+1)} = \sum_{r=1}^{n^2}\frac{1}{(kr+1)(kr-k+1)} - \sum_{r=1}^{n-1}\frac{1}{(kr+1)(kr-k+1)}$ | M1 | Or applies the method of differences again.
$\frac{1}{k}\left(1-\frac{1}{kn^2+1}-\left(1-\frac{1}{k(n-1)+1}\right)\right) = \frac{1}{k}\left(\frac{1}{k(n-1)+1}-\frac{1}{kn^2+1}\right)$ | A1 | OE e.g. $\frac{n^2}{kn^2+1} - \frac{(n-1)}{k(n-1)+1}$
---3
\begin{enumerate}[label=(\alph*)]
\item Use the method of differences to find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { ( \mathrm { kr } + 1 ) ( \mathrm { kr } - \mathrm { k } + 1 ) }$ in terms of $n$ and $k$, where $k$ is a positive constant.
\item Deduce the value of $\sum _ { \mathrm { r } = 1 } ^ { \infty } \frac { 1 } { ( \mathrm { kr } + 1 ) ( \mathrm { kr } - \mathrm { k } + 1 ) }$.
\item Find also $\sum _ { \mathrm { r } = \mathrm { n } } ^ { \mathrm { n } ^ { 2 } } \frac { 1 } { ( \mathrm { kr } + 1 ) ( \mathrm { kr } - \mathrm { k } + 1 ) }$ in terms of $n$ and $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q3 [7]}}