June 2024 Paper 2 Q4
4. A sequence \(u_1, u_2, u_3, \ldots\) is defined by
\[\begin{aligned}u_{n+1} &= ku_n - 5\\ u_1 &= 6\end{aligned}\]where \(k\) is a positive constant.
Given that \(u_3 = -1\)
| Scheme | Marks | AO |
|---|---|---|
| \(u_1 = 6 \Rightarrow u_2 = 6k - 5\) \(u_2 = 6k - 5 \Rightarrow u_3 = k(6k-5) - 5\) \(\Rightarrow k(6k-5) - 5 = -1\) | M1 | 1.1b |
| \(\Rightarrow 6k^2 - 5k - 4 = 0\ *\) | A1* | 2.1 |
| (2) |
Notes
M1: Correct application of the given recurrence relation using \(u_1 = 6\) to find \(u_2\) and then \(u_3\) in terms of \(k\) and sets \(u_3 = -1\)
Condone missing brackets if the intention is clear e.g. \(u_2 = 6k - 5 \Rightarrow u_3 = k\,6k - 5 - 5\)
A1*: Obtains the printed answer with no errors including the “= 0”
This is a given answer so do not condone slips/missing brackets unless they are recovered before the final printed answer.
Alternative:
| Scheme | Marks |
|---|---|
| \(u_3 = -1 \Rightarrow -1 = ku_2 - 5 \Rightarrow u_2 = \dfrac{4}{k}\) \(u_1 = 6 \Rightarrow u_2 = 6k - 5 \Rightarrow \dfrac{4}{k} = 6k - 5\) | M1 |
| \(\Rightarrow 6k^2 - 5k - 4 = 0\ *\) | A1* |
M1: Correct application of the given recurrence relation using \(u_3 = -1\) to find \(u_2\) in terms of \(k\) and then uses \(u_1 = 6\) to find another expression for \(u_2\) in terms of \(k\) and equates the 2 expressions.
A1*: Obtains the printed answer with no errors including the “= 0”
This is a given answer so do not condone slips unless they are recovered before the final printed answer.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(k = \dfrac{4}{3}\) | B1 | 2.2a |
| (ii) \(k = \dfrac{4}{3} \Rightarrow u_2 = \dfrac{4}{3} \times 6 - 5 \Rightarrow \displaystyle\sum_{r=1}^{3} u_r = 6 + \frac{4}{3} \times 6 - 5 - 1\) | M1 | 1.1b |
| \(\displaystyle\sum_{r=1}^{3} u_r = 8\) | A1 | 1.1b |
| (3) | ||
| (5 marks) |
Notes
(b)(i)
B1: Deduces the correct value of \(k\). Ignore any working and just look for this value.
Allow equivalent exact values e.g. \(1\frac{1}{3}\) or \(1.\dot{3}\) but not clearly rounded e.g. 1.333
It must be clear that \(k = \dfrac{4}{3}\) is selected so if both roots are offered score B0 unless \(k = \dfrac{4}{3}\) is clearly intended by the calculation in part (ii)
(ii)
M1: Attempts the second term by e.g. \((\text{their } k) \times 6 - 5\) and then adds 6 and \(-1\) to their second term. E.g. \(6 + \text{``}\dfrac{4}{3}\text{''} \times 6 - 5 - 1\)
If they use \(u_1\) and \(u_3\) they must be as given in the question but condone a clear mis-copy of their \(u_2\) value.
The attempt at the second term may be implied by their value.
Note that they may use \(u_3 = -1\) to find \(u_2\) e.g. \(-1 = \text{``}\dfrac{4}{3}\text{''}u_2 - 5 \Rightarrow u_2 = \text{``}\dfrac{3}{4}\text{''}(5-1) = 3\)
Condone slips when rearranging as long as the intention is clear.
The attempt at the second term may be seen embedded in their attempt at the sum e.g.
\(\displaystyle\sum_{r=1}^{3} u_r = 6 + \underline{\frac{4}{3} \times 6 - 5} - 1\) or e.g. \(\displaystyle\sum_{r=1}^{3} u_r = 6 + \underline{\frac{3}{4}(5-1)} - 1\)
If they use both of their values for \(k\) allow M1.
Alternatives:
Note that \(\displaystyle\sum_{r=1}^{3} u_r = 6 + 6k - 5 - 1 = 6k\) so you may just see an attempt at \(6k\) with their \(\dfrac{4}{3}\).
Note that \(\displaystyle\sum_{r=1}^{3} u_r = 6k^2 + k - 4\) so you may just see an attempt at \(6k^2 + k - 4\) with their \(\dfrac{4}{3}\).
A1: Correct value of 8 and no other values unless rejected.
Correct answer with no working scores both marks.
Allow recovery from an inexact value from part (i) e.g. 1.333