AS June 2023 Q4
4. A sequence \(\{u_n\}\), where \(n \geqslant 0\), satisfies the recurrence relation
\[u_{n+1} = \frac{3}{2}u_n - 2n^2 - 4 \qquad u_0 = k\]where \(k\) is an integer.
Given that \(u_{10} \gt 5000\)
| Scheme | Marks | AO |
|---|---|---|
| (aux equation \(m - \dfrac{3}{2} = 0 \Rightarrow\)) complementary function is \(A(1.5)^n\) | B1 | 2.1 |
| Particular solution try \(u_n = \alpha n^2 + \beta n + \gamma\) and substitute into recurrence relation | M1 | 1.1b |
| \(2\alpha n^2 + (4\alpha + 2\beta)n + (2\alpha + 2\beta + 2\gamma) = (3\alpha - 4)n^2 + 3\beta n + (3\gamma - 8)\) and by comparing quadratic, linear and constant terms gives \(2\alpha = 3\alpha - 4\) \(4\alpha + 2\beta = 3\beta\) \(2\alpha + 2\beta + 2\gamma = 3\gamma - 8\) | dM1 | 1.1b |
| \(\alpha = 4,\ \beta = 16,\ \gamma = 48\) | A1 | 1.1b |
| \((u_n =)\,A(1.5)^n + 4n^2 + 16n + 48\) | ||
| \(u_0 = k \Rightarrow A + 48 = k\) | ddM1 | 3.4 |
| \((u_n =)\,(k - 48)(1.5)^n + 4n^2 + 16n + 48\) | A1 | 1.1b |
| (6) |
Notes
B1: cao (or equivalent e.g. \(A(1.5)^{n-1}\))
M1: correct form for particular solution e.g. \(\alpha n^2 + \beta n + \gamma,\ \alpha(n-1)^2 + \beta(n-1) + \gamma,\) etc. (so anything that is equivalent to a three term quadratic in \(n\) with unknown coefficients and constant term) together with a valid substitution into recurrence relation (so not substituting \(u_n\) on both sides of the recurrence relation)
dM1: compares coefficients and setting up all three equations in \(\alpha, \beta, \gamma\) - dependent on previous M mark
A1: cao for the values of \(\alpha, \beta, \gamma\)
ddM1: use correct initial condition correctly to form a linear equation in their \(A\) and \(k\) – dependent on the two previous M marks
A1: correct particular solution (in terms of \(k\)) – need not see \(u_n =\) but if seen then it must be correct (and therefore \(u_{n+1} =\) is A0)
| Scheme | Marks | AO |
|---|---|---|
| \((k - 48)(1.5)^{10} + 4(10)^2 + 16(10) + 48 \gt 5000\) | M1 | 1.1b |
| \(k \gt 124.163\ldots \Rightarrow k = 125\) | A1 | 2.2a |
| (2) | ||
| (8 marks) |
Notes
M1: dependent on all M marks in (a) – substituting \(n = 10\) and setting the particular solution > 5 000 (or equal to)
A1: cao (125) from correct working so must have had a correct expression for \(u_n\) in (a) (so dependent on at least the first 5 marks in (a))
Additional guidance:
Those candidates who re-write \(u_{n+1} = \frac{3}{2}u_n - 2n^2 - 4\) as \(u_n = \frac{3}{2}u_{n-1} - 2(n-1)^2 - 4\) can score full marks. Their solution will usually begin: CF is \(A(1.5)^n\) then a PS of the form \(\alpha n^2 + \beta n + \gamma\) leading to
\[\alpha n^2 + \beta n + \gamma = \left(\frac{3}{2}\alpha - 2\right)n^2 + \left(-3\alpha + \frac{3}{2}\beta + 4\right)n + \left(\frac{3}{2}\alpha - \frac{3}{2}\beta + \frac{3}{2}\gamma - 6\right)\]and then \(\frac{3}{2}\alpha - 2 = \alpha \Rightarrow \alpha = 4,\ -3\alpha + \frac{3}{2}\beta + 4 = \beta \Rightarrow \beta = 16,\ \frac{3}{2}\alpha - \frac{3}{2}\beta + \frac{3}{2}\gamma - 6 = \gamma \Rightarrow \gamma = 48\) and then their general solution should be as in the main scheme (although for the final mark in (a) do look out for those who call the left-hand side \(u_{n+1}\))
It is common for candidates to re-write \(u_{n+1} = \frac{3}{2}u_n - 2n^2 - 4\) as \(u_n = \frac{3}{2}u_{n-1} - 2n^2 - 4\) which is incorrect. This can score all B and M marks in both parts only.
Slightly less common is to have a CF of the form \(A(1.5)^{n-1}\) and a PS of the form \(\alpha(n-1)^2 + \beta(n-1) + \gamma\) this leads to \(u_{n+1} = A(1.5)^{n-1} + 4(n-1)^2 + 32(n-1) + 96\)
Now to use the initial condition correctly \(u_0 = k \Rightarrow u_1 = \frac{3}{2}k - 4\) (oe) and this leads to \(A = \frac{9}{4}k - 108\)
So, \(u_{n+1} = \left(\frac{9}{4}k - 108\right)(1.5)^{n-1} + 4(n-1)^2 + 32(n-1) + 96\) which when re-written in terms of \(u_n\) gives the form as in the main mark scheme
Of course, any CF of the form \(A(1.5)^{n \pm k_1}\) and any PS of the form \(a(n \pm k_2)^2 + b(n \pm k_3) + c\) where \(k_1, k_2, k_3\) are constants will work. So, award the M marks for the correct methods as illustrated in the notes in mark scheme and the first A mark in (a) for three correct coefficients of their PS and the second A mark in (a) for a fully correct expression