| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with exponential terms |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on method of differences with exponential terms. Part (a) requires recognizing the factorization (e^x - 1)^2 and applying telescoping series techniques—standard for FM students. Part (b) tests geometric series convergence (routine). Part (c) applies logarithm properties and standard summation formulae. While multi-step, each component follows established FM techniques without requiring novel insight, making it moderately above average difficulty. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^{3x} - 2e^{2x} + e^x\) | M1 A1 | Shows enough complete terms to make pattern of cancelling clear. GP method: \((e^{2x} - 2e^x + 1)\sum_{r=1}^{n} e^{rx} = (e^{2x} - 2e^x + 1)e^x \frac{(e^x)^n - 1}{e^x - 1}\) |
| \(= e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}\) | A1 | OE: \(e^x(e^x-1)(e^{nx}-1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x < 0\) | B1 | Accept \(x \leq 0\) |
| \(e^{nx} \to 0\) as \(n \to \infty\) leading to \(u_1 + u_2 + u_3 + \ldots = e^x - e^{2x}\) | M1 A1 | Must see \(e^{nx} \to 0\) [as \(n \to \infty\)] agreeing with their set of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n}(rx + \ln(e^x - 1)^2)\) | M1\* | Uses laws of logarithms correctly |
| \(\sum_{r=1}^{n} \ln u_r = \frac{1}{2}xn(n+1) + n\ln(e^x-1)^2\) | dM1 A1 | Applies \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\). AEF for \(+n\ln(e^x-1)^2\) |
## Question 4(a):
$\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^{3x} - 2e^{2x} + e^x$ | **M1 A1** | Shows enough complete terms to make pattern of cancelling clear. GP method: $(e^{2x} - 2e^x + 1)\sum_{r=1}^{n} e^{rx} = (e^{2x} - 2e^x + 1)e^x \frac{(e^x)^n - 1}{e^x - 1}$
$= e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}$ | **A1** | OE: $e^x(e^x-1)(e^{nx}-1)$
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## Question 4(b):
$x < 0$ | **B1** | Accept $x \leq 0$
$e^{nx} \to 0$ as $n \to \infty$ leading to $u_1 + u_2 + u_3 + \ldots = e^x - e^{2x}$ | **M1 A1** | Must see $e^{nx} \to 0$ [as $n \to \infty$] agreeing with their set of $x$
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## Question 4(c):
$\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n}(rx + \ln(e^x - 1)^2)$ | **M1\*** | Uses laws of logarithms correctly
$\sum_{r=1}^{n} \ln u_r = \frac{1}{2}xn(n+1) + n\ln(e^x-1)^2$ | **dM1 A1** | Applies $\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$. AEF for $+n\ln(e^x-1)^2$
---4 Let $\mathrm { u } _ { \mathrm { r } } = \mathrm { e } ^ { \mathrm { rx } } \left( \mathrm { e } ^ { 2 \mathrm { x } } - 2 \mathrm { e } ^ { \mathrm { x } } + 1 \right)$.
\begin{enumerate}[label=(\alph*)]
\item Using the method of differences, or otherwise, find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \mathrm { u } _ { \mathrm { r } }$ in terms of $n$ and $x$.
\item Deduce the set of non-zero values of $x$ for which the infinite series
$$u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$$
is convergent and give the sum to infinity when this exists.
\item Using a standard result from the list of formulae (MF19), find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \ln \mathrm { u } _ { \mathrm { r } }$ in terms of $n$ and $x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q4 [9]}}