AS June 2022 Paper 1 Q9
9
(a) Show that, for \(r \gt 0\),\[\ln(r + 2) - \ln r = \ln\left(1 + \frac{2}{r}\right)\] [1 mark]
(b) 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. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Completes a rigorous argument to show that \(\ln(r + 2) - \ln r = \ln\left(1 + \dfrac{2}{r}\right)\) | R1 | 2.1 |
| (1) |
Typical solution
\[\begin{aligned}\ln(r + 2) - \ln r &= \ln\left(\frac{r + 2}{r}\right) \\ &= \ln\left(1 + \frac{2}{r}\right)\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Writes at least two pairs of logs in the form \(\ln(r + 2) - \ln r\) | M1 | 1.1a |
| Writes at least three pairs of logs in the form \(\ln(r + 2) - \ln r\) including the first pair, the last pair, and at least one other pair. | M1 | 1.1a |
| Correctly reduces the expression to three or four log terms. Condone missing brackets. | A1 | 1.1b |
| Completes a fully correct proof to reach the required result. This mark is only available if all previous marks have been awarded. Must include at least one pair of cancelling terms. Must include correct use of brackets throughout. | R1 | 2.1 |
| (4) | ||
| (5 marks) |
Typical solution
\[\begin{aligned}\sum_{r=1}^{n} \ln\left(1 + \frac{2}{r}\right) &= \sum_{r=1}^{n}\big(\ln(r + 2) - \ln r\big) \\ &= \ln 3 - \ln 1 \\ &\quad + \ln 4 - \ln 2 \\ &\quad + \ln 5 - \ln 3 \\ &\quad + \ldots\ldots\ldots \\ &\quad + \ln n - \ln(n - 2) \\ &\quad + \ln(n + 1) - \ln(n - 1) \\ &\quad + \ln(n + 2) - \ln n \\ &= \ln(n + 2) + \ln(n + 1) - \ln 2 - \ln 1 \\ &= \ln\left(\frac{1}{2}(n + 1)(n + 2)\right)\end{aligned}\]Notes
(corrected from the printed mark scheme: the first line of the typical solution prints \(\displaystyle\sum_{r=1}^{n}\left(1 + \frac{2}{r}\right)\), with the \(\ln\) missing.)