A2 June 2024 Q2
2. The general solution of the first order recurrence relation
\[u_{n+1} + au_n = bn^2 + cn + d \qquad n \geqslant 0\]is given by
\[u_n = A(3)^n + 5n^2 + 1\]where \(A\) is an arbitrary non-zero constant.
By considering expressions for \(u_{n+1}\) and \(u_n\), find the values of the constants \(a\), \(b\), \(c\) and \(d\). (3)
| Scheme | Marks | AO |
|---|---|---|
| \(u_{n+1} = A(3)^{n+1} + 5(n+1)^2 + 1\) | B1 | 1.1b |
| \(u_{n+1} = A(3)^n(3) + 5(n+1)^2 + 1\) \(\Rightarrow u_{n+1} = 3(u_n - 5n^2 - 1) + 5(n+1)^2 + 1\) | M1 | 1.1b |
| \(u_{n+1} - 3u_n = -10n^2 + 10n + 3\) | A1 | 1.1b |
| (3) | ||
| (3 marks) |
Notes
B1: any correct expression for \(u_{n+1}\)
M1: eliminating \(A\) to form a first order recurrence relation containing \(u_{n+1}\) and \(u_n\)
A1: CAO for \(u_{n+1} - 3u_n = -10n^2 + 10n + 3\) (need not explicitly state \(a = -3, b = -10, c = 10, d = 3\))
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| \(u_{n+1} + au_n = 0\) \(u_{n+1} = -au_n\) C.F. \(u_n = A(-a)^n\) \(-a = 3 \quad \Rightarrow \quad a = -3\) | B1 | |
| Particular solution Try \(\lambda n^2 + \mu n + \nu \quad \Rightarrow \quad \lambda = 5 \quad \mu = 0 \quad \nu = 1\) \(5(n+1)^2 + 1 - 3\left(5n^2 + 1\right) = bn^2 + cn + d\) \(\Rightarrow b = -10 \quad c = 10 \quad d = 3\) | M1 A1 | |
| (3) |
B1: Considers the C.F. and deduces \(a = -3\) with no other values stated
M1: Obtains P.S. and forms equation using \(u_{n+1} \pm 3u_n\)
A1: CAO for \(u_{n+1} - 3u_n = -10n^2 + 10n + 3\) (need not explicitly state \(a = -3, b = -10, c = 10, d = 3\))
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| \(u_{n+1} + au_n = 3A(3)^n + 5n^2 + 10n + 6\) \(\qquad + aA(3)^n + 5an^2 + a\) | B1 | |
| Compares coefficients \(3 + a = 0 \quad \Rightarrow \quad a = -3\) \(5 + 5a = b \quad \Rightarrow \quad b = -10\) \(c = 10\) \(6 + a = d \quad \Rightarrow \quad d = 3\) | M1 A1 | |
| (3) |
B1: Forms the correct equation for \(u_{n+1} + au_n\)
M1: Attempts to compare coefficients – at least three terms seen
A1: CAO