A2 June 2024 Paper 2 Q1
1
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\displaystyle\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r + 1}\right) =\right) \dfrac{1}{1} - \cancel{\dfrac{1}{2}} + \cancel{\dfrac{1}{2}} - \cancel{\dfrac{1}{3}} + \cancel{\dfrac{1}{3}} - \cancel{\dfrac{1}{4}} + \ldots\) \(\left(\ldots + \cancel{\dfrac{1}{n - 1}}\right) - \cancel{\dfrac{1}{n}} + \cancel{\dfrac{1}{n}} - \dfrac{1}{n + 1} = 1 - \dfrac{1}{n + 1}\) | B1 | 1.1 |
| [1] |
Notes
B1: AG. Writing the sum out so that the cancellation pattern is clear. If dots missing then some indication of existence of intermediate terms must be seen. Condone absence of + signs if listed by \(r\) value.
Must see \(r = 1\) and \(r = n\) and at least two justified cancellations.
Or algebraic cancellation: \(\displaystyle = \sum_{r=1}^{n}\frac{1}{r} - \sum_{r=1}^{n}\frac{1}{r + 1} = \sum_{r=1}^{n}\frac{1}{r} - \sum_{r=2}^{n+1}\frac{1}{r} = \frac{1}{1} + \sum_{r=2}^{n}\frac{1}{r} - \left(\sum_{r=2}^{n}\frac{1}{r} + \frac{1}{n + 1}\right) = 1 - \frac{1}{n + 1}\)
(Corrected from the printed mark scheme: the first line is printed with a stray bracket, as \(\ldots + \frac{1}{2} - \left(\frac{1}{3} + \frac{1}{3} - \frac{1}{4}\right) + \ldots\); the terms are \(-\frac{1}{3} + \frac{1}{3} - \frac{1}{4}\), as typed above.)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\left(1 - \dfrac{1}{1 + 99} = 1 - \dfrac{1}{100}\right) = \dfrac{99}{100}\) or 0.99 oe | B1 | 1.1 |
| [1] | ||
| (ii) eg \(\displaystyle\lim_{n \to \infty} \frac{1}{n + 1} = 0\) or \(\displaystyle\lim_{n \to \infty} \sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r + 1}\right) = 1\) | B1 | 1.1 |
| \(\displaystyle\left(\sum_{r=100}^{\infty}\left(\frac{1}{r} - \frac{1}{r + 1}\right)\right) = \sum_{r=1}^{\infty}\left(\frac{1}{r} - \frac{1}{r + 1}\right) - \sum_{r=1}^{99}\left(\frac{1}{r} - \frac{1}{r + 1}\right)\) | M1 | 1.1 |
| \(= 1 - \dfrac{99}{100} = \dfrac{1}{100}\) www | A1 | 1.1 |
| [3] |
Notes
(b)(ii)
B1: For correct consideration of limit as \(n\) tends to infinity. Could be embedded but limit must be explicitly evaluated.
B0 for use of \(\infty\) in fraction (unless correct limit work seen) or incorrect limit notation eg \(\displaystyle\lim_{n \to \infty} \frac{1}{n + 1} \to 0\)
M1: Rewriting sum as difference of two sums with correct limits. Condone missing summation content but correct relationship between summation limits must be clear.
A1: If B0M0 then SCB1 for 0.01 from \(1 - 0.99 = 0.01\).