AS June 2022 Q4
4. A sequence \(\{u_n\}\), where \(n \geqslant 0\), satisfies the recurrence relation
\[u_{n+1} + 3u_n = n + k\]where \(k\) is a non-zero constant.
Given that \(u_0 = 1\)
Given that \(u_n\) is a linear function of \(n\),
| Scheme | Marks | AO |
|---|---|---|
| (aux equation \(m + 3 = 0 \Rightarrow\)) complementary function is \(A(-3)^n\) | B1 | 2.1 |
| Particular solution try \(u_n = an + b\) and substitute into recurrence relation | M1 | 1.1b |
| \(a(n+1) + b + 3(an + b) = n + k\) and by comparing linear and constant terms gives \(a + 3a = 1\) \(a + b + 3b = k\) | dM1 | 1.1b |
| \(a = \dfrac{1}{4},\ 4b + \dfrac{1}{4} = k \Rightarrow b = \dfrac{4k - 1}{16}\) | ddM1 | 1.1b |
| \((u_n =)\,A(-3)^n + \dfrac{1}{4}n + \dfrac{4k-1}{16}\) or \((u_{n+1} =)\,A(-3)^{n-1} + \dfrac{1}{4}(n-1) + \dfrac{4k+7}{16}\) | A1 | 1.1b |
| \(u_0 = 1 \Rightarrow A + \dfrac{4k-1}{16} = 1\) | dddM1 | 3.4 |
| \(u_n = \left(\dfrac{17 - 4k}{16}\right)(-3)^n + \dfrac{1}{4}n + \dfrac{4k-1}{16}\) | A1 | 1.1b |
| (7) |
Notes
B1: cao (or equivalent e.g. \(A(-3)^{n-1}\)) – condone \(A(-3^n)\) in (a) for all but the final A mark
M1: correct form for particular solution e.g. \(an + b,\ a(n-1) + b,\) etc. (so anything that is a constant times \(n\) + a constant) and substituted into recurrence relation
dM1: compares coefficients and setting up both equations in \(a, b, k\) (so one equation in \(a\) only and one equation in \(a\), \(b\) and \(k\) (although \(a\) may have already been found from the first equation)) – dependent on previous M mark
ddM1: solve for \(a\) and \(b\) (with \(b\) in terms of \(k\)) – dependent on both previous M marks
A1: a correct general solution (in terms of \(k\)) – ignore labelling of left-hand side
dddM1: use correct initial condition correctly to form an equation in their \(A\) and \(k\) – dependent on all three previous M marks
A1: correct particular solution (in terms of \(k\)) – must have correct left-hand side
| Scheme | Marks | AO |
|---|---|---|
| Setting \(\dfrac{17 - 4k}{16} = 0\) and solving for \(k\) | M1 | 3.1a |
| \(k = \dfrac{17}{4} \Rightarrow u_{100} = \dfrac{1}{4}(100) + 1\) | dM1 | 1.1b |
| \(u_{100} = 26\) | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
M1: Setting the coefficient of the exponential term equal to zero and solving for \(k\) – their solution from (a) must be of the form \(\alpha(\beta)^n + \gamma n + \delta\) (where \(\alpha, \beta, \gamma, \delta\) are constants and \(\alpha, \delta\) are in terms of \(k\) only)
dM1: Substituting their value of \(k\) to obtain an expression for \(u_n\) which is linear and substituting \(n = 100\) (dependent on previous M mark)
A1: cao (26) – from correct working including a correct expression for \(u_n\) in (a)
Additional guidance:
Those candidates who re-write \(u_{n+1} + 3u_n = n + k\) as \(u_n + 3u_{n-1} = (n-1) + k\) can score full marks. Their solution will usually begin: CF is \(A(-3)^n\) then a PS of the form \(an + b\) leading to \(4an + 4b - 3a - n + 1 - k = 0\) and so \(a = \frac{1}{4}, b = \frac{4k-1}{16}\) 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 \(u_{n+1} + 3u_n = n + k\) as \(u_n + 3u_{n-1} = n + k\) 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 \(A(-3)^{n-1}\) and a PS of the form \(a(n-1) + b\) this leads to
\(a(n-1) + a + 3(a(n-2) + b) = n + k\) so \(4an - 7a + 4b = n + k\) therefore \(a = \frac{1}{4}\) and \(b = \frac{4k+7}{16}\)
So, giving as a general solution \((u_{n+1} =)\,A(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}\)
Now to use the initial condition correctly \(u_0 = 1 \Rightarrow u_1 = k - 3\) (oe) and this leads to \(A = \frac{153 - 36k}{16}\)
So, \(u_{n+1} = \left(\frac{153 - 36k}{16}\right)(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}\) 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 \(A(-3)^{n \pm k_1}\) and any PS of the form \(a(n \pm k_2) + b\) where \(k_1, k_2\) 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 a correct general solution (ignoring the labelling of the left-hand side) and the second A mark in (a) for a fully correct expression with correct left-hand side