A2 June 2024 Paper 1 Q1
1 By expressing \(\dfrac{1}{r + 1} - \dfrac{1}{r + 2}\) as a single fraction, find \(\displaystyle\sum_{r=1}^{n} \frac{1}{(r + 1)(r + 2)}\) in terms of \(n\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{r + 1} - \dfrac{1}{r + 2} = \dfrac{1}{(r + 1)(r + 2)}\) | B1 | 1.1 |
| so \(\displaystyle\sum_{r=1}^{n} \frac{1}{(r + 1)(r + 2)} = \sum_{r=1}^{n}\left[\frac{1}{r + 1} - \frac{1}{r + 2}\right]\) \(= \frac{1}{2} - \frac{1}{3} + \frac{1}{3} \ldots\) | M1* | 2.5 |
| \(\ldots - \frac{1}{n + 2}\) | M1dep | 2.1 |
| \(= \dfrac{1}{2} - \dfrac{1}{n + 2}\) | A1 | 2.2a |
| [4] |
Notes
B1: Denominator may be \(r^2 + 3r + 2\). Cannot be implied.
M1*: Enough correct terms to show cancellation in their series. First term must be correct. Fractions need not be simplified
A1: isw (or \(\frac{n}{2(n + 2)}\))
Alternative method
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{r + 1} - \dfrac{1}{r + 2} = \dfrac{1}{(r + 1)(r + 2)}\) | B1 |
| \(\displaystyle\sum_{r=1}^{n} \frac{1}{r + 1} - \sum_{r=2}^{n+1} \frac{1}{r + 1}\) | M1* |
| \(= \frac{1}{1 + 1} - \frac{1}{(n + 1) + 1}\) | M1dep |
| \(= \dfrac{1}{2} - \dfrac{1}{n + 2}\) | A1 |
| [4] |
B1: Denominator may be \(r^2 + 3r + 2\). Cannot be implied.
M1*: Rewriting so both series have same fraction with correct limits
M1dep: Correct substitution of limits to leave two terms
A1: isw (or \(\frac{n}{2(n + 2)}\))