A2 June 2022 Q6
6.
(a) Determine the general solution of the recurrence relation\[u_n = 2u_{n-1} - u_{n-2} + 2^n \qquad n \geqslant 2\] (4)
(b) Hence solve this recurrence relation given that \(u_0 = 2u_1\) and \(u_4 = 3u_2\) (2)
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda^2 - 2\lambda + 1 = 0 \Rightarrow \lambda = \ldots \{\lambda = 1\}\) | M1 | 1.1b |
| \(u_n = A + Bn\) | A1 | 2.2a |
| \(u_n = \mu(2)^n \Rightarrow \mu(2)^n = 2\mu(2)^{n-1} - \mu(2)^{n-2} + (2)^n\) \(\Rightarrow 4\mu(2)^{n-2} = 4\mu(2)^{n-2} - \mu(2)^{n-2} + 4(2)^{n-2}\) \(\Rightarrow \mu = \ldots \{4\}\) | M1 | 1.1b |
| \(u_n = A + Bn + 4(2)^n\) or \(u_n = A + Bn + (2)^{n+2}\) | A1 | 1.1b |
| (4) |
Notes
M1: Forms and solves the auxiliary equation.
A1: Correct complementary function.
M1: Correct form for the particular solution, substitutes into the recurrence relation to find the PS.
A1: Correct general solution
(corrected from the printed mark scheme: the first line is printed in a garbled symbol font)
| Scheme | Marks | AO |
|---|---|---|
| Uses the information to find the values of the constants For example \(u_0 = 2u_1 \Rightarrow A + 4 = 2(A + B + 4(2)) \Rightarrow \ldots \{A + 2B = -12\}\) \(u_4 = 3u_2 \Rightarrow A + 4B + \text{‘}4\text{’}(2)^4 = 3(A + 2B + 4(2)^2) \Rightarrow \ldots \{2A + 2B = 16\}\) Solves simultaneous equations to find values for \(A\) and \(B\). Alternatively \(u_2 = 2u_1 - u_0 + 2^2 = u_0 - u_0 + 4 = 4\) leading to \(u_4 = 3u_2 = 3 \times 4 = 12\) \(u_3 = 2u_2 - u_1 + 2^3 = 2 \times 4 - \left(\dfrac{A+4}{2}\right) + 8 = 16 - \left(\dfrac{A+4}{2}\right)\) \(u_4 = 2u_3 - u_2 + 2^4 = 2\left(16 - \left(\dfrac{A+4}{2}\right)\right) - 4 + 16 = 12\) Leading \(A = \ldots\ u_2 = \text{‘}28\text{’} + 2B + 4(2^2) = 4 \Rightarrow B = \ldots\) | M1 | 3.1a |
| \(u_n = 28 - 20n + 4(2)^n\) or \(u_n = 28 - 20n + (2)^{n+2}\) | A1 | 1.1b |
| (2) | ||
| (6 marks) |
Notes
Note: They must have two constants to score any marks in this part
M1: A complete method to find the constants using the information given. Form two equations and solves simultaneously to find values for \(A\) and \(B\).
A1: Correct solution.