June 2022 Paper 3 Q3
3 An infinite sequence \(a_1, a_2, a_3, \ldots\) is defined by \(a_n = \dfrac{n}{n+1}\), for all positive integers \(n\).
| Scheme | Marks | AO |
|---|---|---|
| 1 | B1 | 2.2a |
| [1] |
Notes
B1: Cao
Must be clear that the limit is equal to 1. Therefore tending to or approaching 1 or 0.999 or \(\to 1\) etc do not score
| Scheme | Marks | AO |
|---|---|---|
| \(a_n = 1 - \dfrac{1}{n+1}\) | M1 | 3.1a |
| \(n+1\) increases as \(n\) increases so \(\dfrac{1}{n+1}\) decreases | M1 | 2.4 |
| Less is being taken away from 1 each time so \(a_n = 1 - \dfrac{1}{n+1}\) increases | E1 | 2.1 |
| [3] |
Notes
M1: Convincing explanation that \(\dfrac{1}{n+1}\) decreases
E1: Convincing completion (A.G.)
Special Case (Max 2 for this method)
Correct differentiation and states \(> 0\) to demonstrate increasing SC B1
Goes on to consider [positive] integers as a subset of reals to complete their argument SC B1 dep on first B1.
Alternative method 1
| Scheme | Marks |
|---|---|
| \(\dfrac{n+1}{n+2}\) | M1 |
| \(n(n+2) < (n+1)^2\) | M1 |
| \(n(n+2) = n^2 + 2n\) and \((n+1)^2 = n^2 + 2n + 1\) Expanding brackets to give convincing completion (A.G.) | E1 |
M1: This method is testing \(\dfrac{n}{n+1} < \dfrac{n+1}{n+2}\)
Formulating expression for \((n+1)\)th or \((n-1)\)th term
M1: For formulating inequality and cross multiplying
E1: Probably see \(\dfrac{n+1}{n+2} > \dfrac{n}{n+1}\) [i.e. increasing]
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\dfrac{n+1}{n+2}\) | M1 |
| Difference is \(\dfrac{n+1}{n+2} - \dfrac{n}{n+1} = \dfrac{(n+1)^2 - n(n+2)}{(n+1)(n+2)}\) | M1 |
| \(\dfrac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \dfrac{1}{(n+1)(n+2)} > 0\) So sequence is increasing | E1 |
M1: Formulating expression for \((n+1)\)th term
M1: Finding difference and attempts to form a single fraction eg may be \(a_n - a_{n+1}\)
E1: Convincing completion (A.G.)