A2 June 2023 Q6
6. Determine a closed form for the recurrence relation\[u_0 = 1 \qquad u_1 = 4\]\[u_{n+2} = 2u_{n+1} - \frac{4}{3}u_n + n \qquad n \geqslant 0\]
| Scheme | Marks | AO |
|---|---|---|
| \[\lambda^2 = 2\lambda - \frac{4}{3} \Rightarrow 3\lambda^2 - 6\lambda + 4 = 0 \Rightarrow \lambda = \ldots\] | M1 | 1.1b |
| \[\lambda = \frac{6 \pm \sqrt{36 - 48}}{6} = \frac{3 \pm \mathrm{i}\sqrt{3}}{3} = \frac{2}{\sqrt{3}}\left(\frac{\sqrt{3}}{2} \pm \frac{1}{2}\mathrm{i}\right) = \frac{2\sqrt{3}}{3}\mathrm{e}^{\pm\mathrm{i}\frac{\pi}{6}}\] | A1 | 1.1b |
| CF is \[u_n = A\left(\frac{3+\mathrm{i}\sqrt{3}}{3}\right)^n + B\left(\frac{3-\mathrm{i}\sqrt{3}}{3}\right)^n\] o.e. or \[\left(\frac{2\sqrt{3}}{3}\right)^n\left(P\cos\left(\frac{\pi n}{6}\right) + Q\sin\left(\frac{\pi n}{6}\right)\right)\] | A1ft | 2.2a |
| \[u_n = kn + l \Rightarrow k(n + 2) + l = 2k(n + 1) + 2l - \frac{4}{3}(kn + l) + n\]\[\Rightarrow \left(1 - \frac{1}{3}k\right)n - \frac{1}{3}l = 0 \Rightarrow k = \ldots, l = \ldots \quad (k = 3, l = 0)\] | M1 | 1.1b |
| \[u_n = A\left(\frac{3+\mathrm{i}\sqrt{3}}{3}\right)^n + B\left(\frac{3-\mathrm{i}\sqrt{3}}{3}\right)^n + 3n\] o.e. or \[u_n = \left(\frac{2\sqrt{3}}{3}\right)^n\left(P\cos\left(\frac{\pi n}{6}\right) + Q\sin\left(\frac{\pi n}{6}\right)\right) + 3n\] | A1 | 1.1b |
| \[\left.\begin{aligned} &u_0 = 1 \Rightarrow A + B = 1 \\ &u_1 = 4 \Rightarrow A + B + (A - B)\frac{\mathrm{i}\sqrt{3}}{3} = 1 \end{aligned}\right\} \Rightarrow (A - B)\frac{\mathrm{i}\sqrt{3}}{3} = 0 \Rightarrow A = \ldots, B = \ldots\]\[\left.\begin{aligned} &u_0 = 1 \Rightarrow P + 0Q = 1 \\ &u_1 = 4 \Rightarrow \frac{2}{\sqrt{3}}\left(P\frac{\sqrt{3}}{2} + \frac{Q}{2}\right) = 1 \end{aligned}\right\} \Rightarrow P = \ldots \Rightarrow Q = \ldots\]Solves simultaneous equations to find values for \(A\) and \(B\) or \(P\) and \(Q\) | M1 | 3.1a |
| \[\left(A = B = \frac{1}{2} \text{ or } P = 1,\ Q = 0\right)\]\[\Rightarrow u_n = \frac{1}{2}\left(\frac{3+\mathrm{i}\sqrt{3}}{3}\right)^n + \frac{1}{2}\left(\frac{3-\mathrm{i}\sqrt{3}}{3}\right)^n + 3n\] o.e. Or \[u_n = \left(\frac{2\sqrt{3}}{3}\right)^n\cos\left(\frac{\pi n}{6}\right) + 3n\] | A1 | 1.1b |
| (7) | ||
| (7 marks) |
Notes
M1: Forms and solves the auxiliary equation.
A1: Correct roots – either Cartesian or polar form, award when first seen and isw.
A1ft: Correct complementary function, follow through on their first complex roots. (so A1A0 if roots initially correct but error simplifying leads to wrong CF). Note: use of power \(n + 1\) or \(n - 1\) in Cartesian form is also fine
M1: Correct form for the particular solution and a complete method to find the PS.
A1: Correct general solution, either form
M1. Substitutes \(n = 1\) and sets equal to 4 and substitutes \(n = 0\) and sets equal to 1. To find the values of the constants
A1: Correct solution, any form.