| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Partial fractions then method of differences |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring partial fractions, method of differences with telescoping series, and limit manipulation. Part (a) is standard technique, part (b) is routine convergence, but part (c) requires insight to handle the variable limits and factor of n, making it moderately challenging overall for Further Maths. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{r^2-1} = \frac{1}{(r-1)(r+1)} = \frac{1}{2}\left(\frac{1}{r-1} - \frac{1}{r+1}\right)\) | M1 A1 | Partial fractions |
| \(\sum_{r=2}^{n} \frac{1}{r^2-1} = \frac{1}{2}\left(1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \cdots + \frac{1}{n-2} - \frac{1}{n} + \frac{1}{n-1} - \frac{1}{n+1}\right)\) | M1 | Telescoping sum written out |
| \(= \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}\right)\) | A1 | After cancellation |
| \(= \frac{3}{4} - \frac{2n+1}{2n(n+1)}\) | A1 | Simplified form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{3}{4}\) | B1 | Limit as \(n \to \infty\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=n+1}^{2n} \frac{n}{r^2-1} = \sum_{r=2}^{2n} \frac{n}{r^2-1} - \sum_{r=2}^{n} \frac{n}{r^2-1}\) | M1 | Splitting the sum |
| \(= n\!\left(\frac{3}{4} - \frac{4n+1}{4n(2n+1)}\right) - n\!\left(\frac{3}{4} - \frac{2n+1}{2n(n+1)}\right) = \frac{n(2n+1)}{2n(n+1)} - \frac{n(4n+1)}{4n(2n+1)}\) | M1 A1 | Substituting and simplifying |
| \(\to \frac{1}{2}\) as \(n \to \infty\) | A1 | Correct limit |
## Question 4:
### Part 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{r^2-1} = \frac{1}{(r-1)(r+1)} = \frac{1}{2}\left(\frac{1}{r-1} - \frac{1}{r+1}\right)$ | M1 A1 | Partial fractions |
| $\sum_{r=2}^{n} \frac{1}{r^2-1} = \frac{1}{2}\left(1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \cdots + \frac{1}{n-2} - \frac{1}{n} + \frac{1}{n-1} - \frac{1}{n+1}\right)$ | M1 | Telescoping sum written out |
| $= \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}\right)$ | A1 | After cancellation |
| $= \frac{3}{4} - \frac{2n+1}{2n(n+1)}$ | A1 | Simplified form |
### Part 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{3}{4}$ | B1 | Limit as $n \to \infty$ |
### Part 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=n+1}^{2n} \frac{n}{r^2-1} = \sum_{r=2}^{2n} \frac{n}{r^2-1} - \sum_{r=2}^{n} \frac{n}{r^2-1}$ | M1 | Splitting the sum |
| $= n\!\left(\frac{3}{4} - \frac{4n+1}{4n(2n+1)}\right) - n\!\left(\frac{3}{4} - \frac{2n+1}{2n(n+1)}\right) = \frac{n(2n+1)}{2n(n+1)} - \frac{n(4n+1)}{4n(2n+1)}$ | M1 A1 | Substituting and simplifying |
| $\to \frac{1}{2}$ as $n \to \infty$ | A1 | Correct limit |4
\begin{enumerate}[label=(\alph*)]
\item By first expressing $\frac { 1 } { r ^ { 2 } - 1 }$ in partial fractions, show that
$$\sum _ { r = 2 } ^ { n } \frac { 1 } { r ^ { 2 } - 1 } = \frac { 3 } { 4 } - \frac { a n + b } { 2 n ( n + 1 ) }$$
where $a$ and $b$ are integers to be found.
\item Deduce the value of $\sum _ { r = 2 } ^ { \infty } \frac { 1 } { r ^ { 2 } - 1 }$.
\item Find the limit, as $n \rightarrow \infty$, of $\sum _ { r = n + 1 } ^ { 2 n } \frac { n } { r ^ { 2 } - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q4 [10]}}