A2 October 2020 Q4
4. The complementary function for the second order recurrence relation
\[u_{n+2} + \alpha u_{n+1} + \beta u_n = 20(-3)^n \qquad n \geqslant 0\]is given by
\[u_n = A(2)^n + B(-1)^n\]where \(A\) and \(B\) are arbitrary non-zero constants.
(a) Find the value of \(\alpha\) and the value of \(\beta\). (2)
Given that \(2u_0 = u_1\) and \(u_4 = 164\)
(b) find the solution of this second order recurrence relation to obtain an expression for \(u_n\) in terms of \(n\). (6)
| Scheme | Marks | AO |
|---|---|---|
| CF of \(u_n = A(2)^n + B(-1)^n \Rightarrow\) auxiliary equation is \((m - 2)(m + 1) = 0\) | M1 | 3.1a |
| \(m^2 - m - 2 = 0 \Rightarrow \alpha = -1,\ \beta = -2\) | A1 | 2.2a |
| (2) |
Notes
M1: Uses given complementary function to find auxiliary equation corresponding to second-order recurrence relation
A1: cao for both \(\alpha\) and \(\beta\)
| Scheme | Marks | AO |
|---|---|---|
| particular solution try \(u_n = \lambda(-3)^n,\ u_{n+1} = \lambda(-3)^{n+1},\ u_{n+2} = \lambda(-3)^{n+2}\) | M1 | 2.1 |
| \(9\lambda + 3\lambda - 2\lambda = 20\ (\Rightarrow \lambda = 2)\) | M1 | 1.1b |
| \(u_n = A(2)^n + B(-1)^n + 2(-3)^n\) | A1 | 1.1b |
| \(2u_0 = u_1 \Rightarrow 2A + 2B + 4 = 2A - B - 6\) | M1 | 1.1b |
| \(u_4 = 164 \Rightarrow 16A + B + 162 = 164\) | M1 | 1.1b |
| \(A = \dfrac{1}{3},\ B = -\dfrac{10}{3} \Rightarrow u_n = \dfrac{1}{3}(2)^n - \dfrac{10}{3}(-1)^n + 2(-3)^n\) | A1 | 2.2a |
| (6) | ||
| (8 marks) |
Notes
M1: substitute \(u_n = \lambda(-3)^n\) into their second-order recurrence relation
M1: forms linear equation in \(\lambda\) only
A1: correct general solution
M1: use \(2u_0 = u_1\) to form an equation in \(B\) (and possibly \(A\))
M1: use \(u_4 = 164\) to set up a second equation in \(A\) and \(B\)
A1: cao