| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with logarithmic terms |
| Difficulty | Standard +0.8 This is a non-standard method of differences question requiring students to recognize logarithmic telescoping, manipulate the sum algebraically, and express the final result in a specific factored form. While the individual steps use A-level techniques, the combination of logarithm laws with telescoping series and the need to identify integer constants makes this moderately challenging, above typical Further Maths questions but not requiring exceptional insight. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ln(r+2) - \ln r = \ln\left(\frac{r+2}{r}\right) = \ln\left(1+\frac{2}{r}\right)\) | R1 | Completes a rigorous argument to show that \(\ln(r+2) - \ln r = \ln\left(1+\frac{2}{r}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Writes at least two pairs of logs in the form \(\ln(r+2) - \ln r\) | M1 | |
| \(\sum_{r=1}^{n}\left(1+\frac{2}{r}\right) = \sum_{r=1}^{n}(\ln(r+2)-\ln r)\) written with at least three pairs including first, last and one other | M1 | |
| \(= \ln 3 - \ln 1 + \ln 4 - \ln 2 + \ln 5 - \ln 3 + \cdots + \ln n - \ln(n-2) + \ln(n+1) - \ln(n-1) + \ln(n+2) - \ln n\) reduces to three or four log terms | A1 | Condone missing brackets |
| \(= \ln(n+2) + \ln(n+1) - \ln 2 - \ln 1 = \ln\left(\frac{1}{2}(n+1)(n+2)\right)\) | R1 | All previous marks must be awarded; must include at least one pair of cancelling terms; must include correct use of brackets throughout |
## Question 9:
### Part 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\ln(r+2) - \ln r = \ln\left(\frac{r+2}{r}\right) = \ln\left(1+\frac{2}{r}\right)$ | R1 | Completes a rigorous argument to show that $\ln(r+2) - \ln r = \ln\left(1+\frac{2}{r}\right)$ |
### Part 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Writes at least two pairs of logs in the form $\ln(r+2) - \ln r$ | M1 | |
| $\sum_{r=1}^{n}\left(1+\frac{2}{r}\right) = \sum_{r=1}^{n}(\ln(r+2)-\ln r)$ written with at least three pairs including first, last and one other | M1 | |
| $= \ln 3 - \ln 1 + \ln 4 - \ln 2 + \ln 5 - \ln 3 + \cdots + \ln n - \ln(n-2) + \ln(n+1) - \ln(n-1) + \ln(n+2) - \ln n$ reduces to three or four log terms | A1 | Condone missing brackets |
| $= \ln(n+2) + \ln(n+1) - \ln 2 - \ln 1 = \ln\left(\frac{1}{2}(n+1)(n+2)\right)$ | R1 | All previous marks must be awarded; must include at least one pair of cancelling terms; must include correct use of brackets throughout |
---9
\begin{enumerate}[label=(\alph*)]
\item Show that, for $r > 0$,
$$\ln ( r + 2 ) - \ln r = \ln \left( 1 + \frac { 2 } { r } \right)$$
9
\item Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \ln \left( 1 + \frac { 2 } { r } \right) = \ln \left( \frac { 1 } { 2 } ( n + a ) ( n + b ) \right)$$
where $a$ and $b$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2022 Q9 [5]}}