A2 June 2025 Q4
4. A sequence is defined by\[u_{n+2} = 6u_{n+1} - 8u_n \qquad n \geqslant 1\]\[u_1 = 399.6 \qquad u_2 = 798.4\]
(Solutions relying entirely on calculator technology are not acceptable.)
(3)| Scheme | Marks | AO |
|---|---|---|
| \(n = 1: u_1 = 200(2)^1 - \tfrac{1}{10}(4)^1 = 399.6\) \(n = 2: u_2 = 200(2)^2 - \tfrac{1}{10}(4)^2 = 798.4\) (Hence true for \(n = 1\) and \(n = 2\)) | B1 | 2.2a |
| Assume true for some \(n = k\) and \(n = k + 1\), so \(u_k = 200(2)^k - \tfrac{1}{10}(4)^k\) and \(u_{k+1} = 200(2)^{k+1} - \tfrac{1}{10}(4)^{k+1}\) | M1 | 2.4 |
| Then \(u_{k+2} = 6\left[200(2)^{k+1} - \dfrac{1}{10}(4)^{k+1}\right] - 8\left[200(2)^k - \dfrac{1}{10}(4)^k\right]\) | M1 | 1.1b |
| e.g. \(u_{k+2} = \left[3\left(200(2)^{k+2}\right) - 2\left(200(2)^{k+2}\right)\right] + \left[-\tfrac{6}{10}(4)^{k+1} + \tfrac{2}{10}(4)^{k+1}\right]\) \(u_{k+2} = 200(2)^{k+2} - \tfrac{4}{10}(4)^{k+1}\) or \(u_{k+2} = \left[3\left(200(2)^{k+2}\right) - 2\left(200(2)^{k+2}\right)\right] + \left[-\tfrac{3}{20}(4)^{k+2} + \tfrac{1}{20}(4)^{k+2}\right]\) | M1 | 1.1b |
| \(u_{k+2} = 200\left(2^{k+2}\right) - \dfrac{1}{10}\left(4^{k+2}\right)\) | A1 | 2.1 |
| Hence if true for \(n = k\) and \(n = k + 1\) then true for \(n = k + 2\). As also true for \(n = 1\) and \(n = 2\), then true for all \(n \in \mathbb{N}\) by mathematical induction. | A1 | 2.4 |
| (6) |
Notes
B1: Checks the closed form works for \(n = 1\) and \(n = 2\). Allow if they use the recurrence to find \(u_3\) and check for \(n = 2\) and \(n = 3\), but a consecutive pair must be checked.
M1: Makes the inductive assumption. If not explicitly made, accept just stating “\(n = k\) and \(n = k + 1\)” as making the assumption these are true – or implied by use of the relevant formulae, as long as the assumption is made clear in the conclusion. May use e.g. \(n = k - 2\) and \(n = k - 1\) instead and show true for \(n = k\). It must be clear it is the closed forms they are assuming, not a recurrence form.
M1: Substitutes expression for \(n = k\) and \(n = k + 1\) (or equivalents) into the recurrence formula.
M1: Uses algebra in an attempt to achieve the required result
e.g. Uses the coefficients of 6 and 8 to
- write as \(2^n\) terms as \(2^{k+2}\) and simplify.
- write as \(4^n\) terms as \(4^{k+1}\) or \(4^{k+2}\) and simplify.
Note this is a method mark so you may score for the attempt even if some of the working is incorrect as long as the intent to reach the correct form is clear and at least one bit of indec work is correct.
A1: Completes the process correctly to the required form
A1: Correct conclusion made at the end. Depends on all three M’s and the A being gained and an attempt at both \(n = 1\) and \(n = 2\) having been shown true. Must convey the underlined ideas of
- true for \(n = 1\) and \(n = 2\)
- if true for two successive cases, it is also true for the next case
- a suitable conclusion that it is true for all positive \(n\)
though accept equivalent wordings for these.
Note Accept work with \(n\) instead of \(k\) throughout the inductive step.
| Scheme | Marks | AO |
|---|---|---|
| \(u_n = 200(2^n) - \tfrac{1}{10}(4^n) = 200(2^n) - \tfrac{1}{10}(2^n)^2 = 0\) Leading to \(2^n = \ldots\{2000\}\) Alt : reaches \(4^n > 2000 \times 2^n \Rightarrow n\log 4 > \log 2000 + n\log 2\) | M1 | 3.1a |
| Solves \(2^n = 2000 \Rightarrow n = \log_2 2000 = \ldots\{10.96\}\) \(2^n = 2000 \Rightarrow n = \tfrac{\log 2000}{\log 2} = \ldots\{10.96\}\) Alt: Solve the linear equation to find \(n\) | dM1 | 1.1b |
| \(u_{11} = -9830.4\) or \(-\dfrac{49152}{5}\) | A1 | 2.2a |
| (3) | ||
| (9 marks) |
Notes
M1: Sets closed form = 0 or < 0 and solves to set up an inequality, or to find a non-zero value, for \(2^n\). Alternatively if the \(4^n\) is not written in terms of \(2^n\) score for a correct process for taking logs to get a linear equation in \(n\). Be tolerant with incorrect inequalities for the M marks.
dM1: Solves \(2^n = a\) where \(a > 0\) by any valid means. May be by inspection. In the Alt it is for proceeding to a value for \(n\) from the linear equation.
A1: Deduces the first negative term of the sequence