| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Partial Fractions and Telescoping Series |
| Difficulty | Challenging +1.2 This is a multi-part question on telescoping series that requires recognizing the telescoping pattern (straightforward), applying it with logarithmic functions, determining convergence conditions, and using standard summation formulas. Part (a) is routine, part (b) requires understanding of limits and convergence (moderate), and part (c) involves sum of partial sums with factorization. While it tests multiple techniques and requires careful algebraic manipulation, the telescoping structure is explicitly given and the methods are standard for Further Maths students. Slightly above average difficulty due to the multi-step nature and need for careful handling of logarithms and convergence. |
| Spec | 1.04g Sigma notation: for sums of series1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^{f(1)}-x^{f(2)}+x^{f(2)}-x^{f(3)}+\cdots+x^{f(n)}-x^{f(n+1)}\) | M1 | Writes at least three terms, including last |
| \(=x^{f(1)}-x^{f(n+1)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([0<]x\leq 1\) | B1 | Accept \([0<]x<1\); \(\sum_{r=1}^{n}u_r=1-x^{\ln(n+1)}\) |
| \(\sum_{r=1}^{\infty}u_r=1\) [for \(x<1\)] | B1 | Without wrong working |
| \(\sum_{r=1}^{\infty}u_r=0\) for \(x=1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(x^{2\log_x r}=r^2\) | M1 | |
| \(S_n=1-(n+1)^2=-n^2-2n\) | A1 | |
| \(\sum_{n=1}^{N}-n^2-2n=-\frac{1}{6}N(N+1)(2N+1)-N(N+1)\) | M1 | Substitutes formulae from MF19 |
| \(-\frac{1}{6}N(N+1)(2N+7)\) | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^{f(1)}-x^{f(2)}+x^{f(2)}-x^{f(3)}+\cdots+x^{f(n)}-x^{f(n+1)}$ | M1 | Writes at least three terms, including last |
| $=x^{f(1)}-x^{f(n+1)}$ | A1 | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[0<]x\leq 1$ | B1 | Accept $[0<]x<1$; $\sum_{r=1}^{n}u_r=1-x^{\ln(n+1)}$ |
| $\sum_{r=1}^{\infty}u_r=1$ [for $x<1$] | B1 | Without wrong working |
| $\sum_{r=1}^{\infty}u_r=0$ for $x=1$ | B1 | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $x^{2\log_x r}=r^2$ | M1 | |
| $S_n=1-(n+1)^2=-n^2-2n$ | A1 | |
| $\sum_{n=1}^{N}-n^2-2n=-\frac{1}{6}N(N+1)(2N+1)-N(N+1)$ | M1 | Substitutes formulae from MF19 |
| $-\frac{1}{6}N(N+1)(2N+7)$ | A1 | |5 It is given that $S _ { n } = \sum _ { r = 1 } ^ { n } u _ { r }$, where $u _ { r } = x ^ { \mathrm { f } ( r ) } - x ^ { \mathrm { f } ( r + 1 ) }$ and $x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Find $S _ { n }$ in terms of $n , x$ and the function f .
\item Given that $\mathrm { f } ( r ) = \ln r$, find the set of 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.\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-10_2716_31_106_2016}\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-11_2723_35_101_20}
\item Given instead that $\mathrm { f } ( r ) = 2 \log _ { x } r$ where $x \neq 1$, use standard results from the List of formulae (MF19) to find $\sum _ { n = 1 } ^ { N } S _ { n }$ in terms of $N$. Fully factorise your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q5 [9]}}