A2 June 2022 Q2
2. The general solution of the second order recurrence relation
\[u_{n+2} + k_1u_{n+1} + k_2u_n = 0 \qquad n \geqslant 0\]is given by
\[u_n = (A + Bn)(-3)^n\]where \(A\) and \(B\) are arbitrary non-zero constants.
(a) Find the value of \(k_1\) and the value of \(k_2\) (2)
Given that \(u_0 = u_1 = 1\)
(b) find the value of \(A\) and the value of \(B\). (2)
| Scheme | Marks | AO |
|---|---|---|
| \((m + 3)^2 = 0\) | M1 | 3.1a |
| \(k_1 = 6\) and \(k_2 = 9\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct auxiliary equation (may be implied by 6, 9 correct) from \(u_n = (A + Bn)(-3)^n\)
A1: CAO – allow values stated implicitly i.e., \(u_{n+2} + 6u_{n+1} + 9u_n = 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(u_0 = 1 \Rightarrow A(-3)^0 = 1\) \(u_1 = 1 \Rightarrow (A + B)(-3) = 1\) | M1 | 1.1b |
| \(A = 1\) and \(B = -\dfrac{4}{3}\) | A1 | 1.1b |
| (2) | ||
| (4 marks) |
Notes
M1: Uses \(u_0 = u_1 = 1\) and attempts to find \(A\) and \(B\)
A1: CAO – allow values stated implicitly e.g. \(u_n = \left(1 - \dfrac{4}{3}n\right)(-3)^n\)