| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.8 Part (a) is routine application of standard formulae. Part (b) requires algebraic verification of a given identity, which is straightforward. The method of differences application is standard once the identity is verified. Part (c) requires taking a limit, which is conceptually simple here. This is a typical Further Maths question testing method of differences with multiple parts, but the identity is provided and the steps are guided, making it moderately challenging but not requiring novel insight. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}n(n+1)(2n+1) + \frac{3}{2}n(n+1) + n\) | M1 A1 | Substitutes correct formulae from MF19 |
| \(n^3 + 3n^2 + 3n\) | A1 | Simplifies |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3(r+1)^3} = \frac{r^3 + 3r^2 + 3r + 1 - r^3}{r^3(r+1)^3} = \frac{3r^2+3r+1}{r^3(r+1)^3}\) | M1 A1 | Puts over a common denominator and expands, AG |
| \(\sum_{r=1}^{n} \frac{3r^2+3r+1}{r^3(r+1)^3} = \sum_{r=1}^{n}\left(\frac{1}{r^3} - \frac{1}{(r+1)^3}\right) = 1 - \frac{1}{2^3} + \frac{1}{2^3} - \frac{1}{3^3} + \ldots + \frac{1}{n^3} - \frac{1}{(n+1)^3}\) | M1 A1 | Shows three complete terms, including last |
| \(1 - \frac{1}{(n+1)^3}\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1\) | B1 FT | FT from *their* answer to part (b) |
| Total: 1 |
## Question 1:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}n(n+1)(2n+1) + \frac{3}{2}n(n+1) + n$ | M1 A1 | Substitutes correct formulae from MF19 |
| $n^3 + 3n^2 + 3n$ | A1 | Simplifies |
| **Total: 3** | | |
---
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3(r+1)^3} = \frac{r^3 + 3r^2 + 3r + 1 - r^3}{r^3(r+1)^3} = \frac{3r^2+3r+1}{r^3(r+1)^3}$ | M1 A1 | Puts over a common denominator and expands, AG |
| $\sum_{r=1}^{n} \frac{3r^2+3r+1}{r^3(r+1)^3} = \sum_{r=1}^{n}\left(\frac{1}{r^3} - \frac{1}{(r+1)^3}\right) = 1 - \frac{1}{2^3} + \frac{1}{2^3} - \frac{1}{3^3} + \ldots + \frac{1}{n^3} - \frac{1}{(n+1)^3}$ | M1 A1 | Shows three complete terms, including last |
| $1 - \frac{1}{(n+1)^3}$ | A1 | |
| **Total: 5** | | |
---
**Part (c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1$ | B1 FT | FT from *their* answer to part (b) |
| **Total: 1** | | |1
\begin{enumerate}[label=(\alph*)]
\item Use standard results from the list of formulae (MF19) to find $\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } + 3 r + 1 \right)$ in terms of $n$, simplifying your answer.
\item Show that
$$\frac { 1 } { r ^ { 3 } } - \frac { 1 } { ( r + 1 ) ^ { 3 } } = \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }$$
and hence use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }$.
\item Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q1 [9]}}