| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| 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 This is a Further Maths question requiring partial fractions decomposition (though not explicitly stated), method of differences application with a parameter, telescoping series manipulation, and then solving for a parameter using the limit as nāā. While methodical, it requires multiple techniques and careful algebraic manipulation with the parameter a throughout, placing it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{ar+1} - \dfrac{1}{ar+a+1}\right)\) | M1 A1 | Finds partial fractions |
| \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{a+1} - \dfrac{1}{2a+1} + \dfrac{1}{2a+1} - \dfrac{1}{3a+1} + \ldots + \dfrac{1}{an+1} - \dfrac{1}{an+a+1}\right)\) | M1 | Writes at least three complete terms, including first and last |
| \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{a+1} - \dfrac{1}{an+a+1}\right)\) | A1 | OE ISW |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\sum_{r=1}^{\infty} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a^2+a} = \dfrac{1}{6}\) leading to \(a^2 + a - 6 = 0\) | M1 A1 | Finds sum to infinity, forms quadratic |
| \(a = 2\) | A1 | CAO |
| 3 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{ar+1} - \dfrac{1}{ar+a+1}\right)$ | M1 A1 | Finds partial fractions |
| $\displaystyle\sum_{r=1}^{n} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{a+1} - \dfrac{1}{2a+1} + \dfrac{1}{2a+1} - \dfrac{1}{3a+1} + \ldots + \dfrac{1}{an+1} - \dfrac{1}{an+a+1}\right)$ | M1 | Writes at least three complete terms, including first and last |
| $\displaystyle\sum_{r=1}^{n} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a}\left(\dfrac{1}{a+1} - \dfrac{1}{an+a+1}\right)$ | A1 | OE ISW |
| | **4** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\sum_{r=1}^{\infty} \dfrac{1}{(ar+1)(ar+a+1)} = \dfrac{1}{a^2+a} = \dfrac{1}{6}$ leading to $a^2 + a - 6 = 0$ | M1 A1 | Finds sum to infinity, forms quadratic |
| $a = 2$ | A1 | CAO |
| | **3** | |1 Let $a$ be a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Use the method of differences to find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { ( \mathrm { ar } + 1 ) ( \mathrm { ar } + \mathrm { a } + 1 ) }$ in terms of $n$ and $a$.
\item Find the value of $a$ for which $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( a r + 1 ) ( a r + a + 1 ) } = \frac { 1 } { 6 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q1 [7]}}