AS October 2020 Q4
4. A sequence \(\{u_n\}\), where \(n \geqslant 1\), satisfies the recurrence relation
\[2u_n = u_{n-1} - kn^2 \quad \text{where} \quad 4u_2 - u_0 = 27k^2\]and \(k\) is a non-zero constant.
Show that, as \(n\) becomes large, \(u_n\) can be approximated by a quadratic function of the form \(an^2 + bn + c\) where \(a\), \(b\) and \(c\) are constants to be determined. (8)
| Scheme | Marks | AO |
|---|---|---|
| (aux equation \(2m - 1 = 0 \Rightarrow\)) complementary function is \(A\left(\frac{1}{2}\right)^n\) | B1 | 2.1 |
| Particular solution try \(u_n = \lambda n^2 + \beta n + \alpha\) and substitute into recurrence relation | M1 | 1.1b |
| \(2\lambda n^2 + 2\beta n + 2\alpha = (\lambda - k)n^2 + (-2\lambda + \beta)n + (\lambda - \beta + \alpha)\) \(\Rightarrow 2\lambda = \lambda - k\) \(\quad\ \ 2\beta = -2\lambda + \beta\) \(\quad\ \ 2\alpha = \lambda - \beta + \alpha\) | M1 | 1.1b |
| \(u_n = A\left(\frac{1}{2}\right)^n - kn^2 + 2kn - 3k\) | A1 | 1.1b |
| \(u_0 = A - 3k,\ u_2 = \frac{1}{4}A - 3k \Rightarrow 4\left(\frac{1}{4}A - 3k\right) - (A - 3k) = 27k^2\) | M1 | 3.1a |
| \(27k^2 + 9k = 0 \Rightarrow k = -\frac{1}{3}\quad (k \neq 0)\) | A1ft | 1.1b |
| As \(n\) becomes large \(A\left(\frac{1}{2}\right)^n \to 0\) | B1 | 2.4 |
| \(u_n \to \frac{1}{3}n^2 - \frac{2}{3}n + 1 \quad \left(a = \frac{1}{3}, b = -\frac{2}{3}, c = 1\right)\) | A1 | 2.2a |
| (8 marks) |
Notes
B1: cao
M1: correct form for the particular solution and substituted into recurrence relation
M1: compares coefficients and setting up all three equations in \(\lambda, \beta, \alpha\)
A1: correct general solution (or with consistent value of \(k\))
M1: use initial condition to obtain a quadratic equation in \(k\)
A1ft: correct solution for \(k\) following through their general solution
B1: correct explanation that the exponential term tends to zero as \(n\) becomes large
A1: cao
Alternative for third M mark: Note that candidates may calculate \(k\) immediately by eliminating \(u_1\) from \(2u_1 = u_0 - k\) and \(2u_2 = u_1 - 4k\) and comparing with \(4u_2 - u_0 = 27k^2\) to obtain a quadratic in \(k\)