C1 June 2017 Q3
3. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned} a_1 &= 1 \\ a_{n+1} &= \frac{k(a_n + 1)}{a_n}, \qquad n \geqslant 1 \end{aligned}\]where \(k\) is a positive constant.
Given that \(\displaystyle\sum_{r=1}^{3} a_r = 10\)
| Scheme | Marks |
|---|---|
| \(\left(a_2 =\right)2k\) | B1 |
| \(\left(a_3 =\right)\dfrac{k(\text{"}2k\text{"} + 1)}{\text{"}2k\text{"}}\) | M1 |
| \(\left(a_3 =\right)\dfrac{2k + 1}{2}\) | A1 |
| (3) |
Notes
B1: \(2k\) only
M1: For substituting their \(a_2\) into \(a_3 = \dfrac{k(a_2 + 1)}{a_2}\) to find \(a_3\) in terms of just \(k\)
A1: \(\left(a_3 =\right)\dfrac{2k + 1}{2}\) or exact simplified equivalent such as \(\left(a_3 =\right)k + \dfrac{1}{2}\) or \(\dfrac{1}{2}(2k + 1)\) but not \(k + \dfrac{k}{2k}\) Must be seen in (a) but isw once a correct simplified answer is seen.
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{3} a_r = 10 \Rightarrow 1 + \text{"}2k\text{"} + \text{"}\frac{2k + 1}{2}\text{"} = 10\) | M1 |
| \(\Rightarrow 2 + 4k + 2k + 1 = 20 \Rightarrow k = \ldots\) or e.g. \(\Rightarrow 6k^2 - 17k = 0 \Rightarrow k = \ldots\) | M1 |
| \(\left(k =\right)\dfrac{17}{6}\) | A1 |
| (3) | |
| (6 marks) |
Notes
Note that there are no marks in (b) for using an AP (or GP) sum formula unless their terms do form an AP (or GP).M1: Writes 1 + their \(a_2\) + their \(a_3\) = 10. E.g. \(1 + 2k + \dfrac{2k^2 + k}{2k} = 10\). Must be a correct follow through equation in terms of \(k\) only.
M1: Solves their equation in \(k\) which has come from the sum of 3 terms = 10, and reaches \(k = \ldots\) Condone poor algebra but if a quadratic is obtained then the usual rules apply for solving – see General Principles. (Note that it does not need to be a 3-term quadratic in this case)
A1: \(k = \dfrac{17}{6}\) or exact equivalent e.g. \(2\dfrac{5}{6}\)
Do not allow \(k = \dfrac{8.5}{3}\) or \(k = \dfrac{17/2}{3}\)
Ignore any reference to \(k = 0\).
Allow 2.83 recurring as long as the recurring is clearly indicated e.g. a dot over the 3.