June 2023 Paper 3 Q15
15
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “Approximating series” are reproduced below; the line numbers are those printed on the Insert.
Line 34
This simplifies to \(\displaystyle\sum_{r=1}^{n}\frac{1}{r} \approx \ln n + \frac{13}{24} + \frac{6n+5}{12n(n+1)}\).
The expression given in line 34 is used to calculate \(\displaystyle\sum_{r=1}^{6}\frac{1}{r}\).
Show that the error in the result is less than 1.5% of the true value. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{6}\frac{1}{r} = \frac{49}{20}\) | B1 | 1.1 |
| \(\ln 6 + \dfrac{13}{24} + \dfrac{36 + 5}{12 \times 6 \times 7} = 2.41477..\) % error = \(100 \times \dfrac{0.0352..}{2.45} = 1.438\%\) This is less than 1.5% OR 1.5 % of actual and compare with 0.03523 | B1 | 2.2a |
| [2] |
Notes
B1: 2.45 (need not have working)
May be done BC
B1: Convincing completion