A2 June 2024 Q8
8. A sequence \(\{u_n\}\), where \(n \geqslant 0\), satisfies the recurrence relation
\[2u_{n+2} + 5u_{n+1} = 3u_n + 8n + 2\]A particular solution of this recurrence relation has \(u_0 = 1\) and \(u_1 = k\), where \(k\) is a positive constant. All terms of the sequence are positive.
| Scheme | Marks | AO |
|---|---|---|
| aux. equation is \(2m^2 + 5m - 3 = 0 \Rightarrow m = \ldots\) | M1 | 2.1 |
| complementary function is \(A(0.5)^n + B(-3)^n\) | A1 | 1.1b |
| Particular solution try \(u_n = an + b\) and substitute into recurrence relation | M1 | 1.1b |
| \(2(a(n+2) + b) + 5(a(n+1) + b) = 3(an + b) + 8n + 2\) and by comparing linear and constant terms gives \(4a = 8\) \(9a + 4b = 2\) | dM1 | 1.1b |
| \((u_n =)\ A(0.5)^n + B(-3)^n + 2n - 4\) | A1ft | 1.1b |
| (5) |
Notes
Note mark (a) and (b) together
M1: correct auxiliary equation and attempt to solve (leading to two distinct values of \(m\))
A1: CAO for complementary function (accept if seen in subsequent working)
M1: correct form for particular solution and substituted \(n + 2,\ n + 1,\ n\) into recurrence relation
dM1: compares coefficients and setting up both equations in \(a, b\) (so one equation in \(a\) only and one equation in \(a\) and \(b\) (although \(a\) may have already been found from the first equation)) – dependent on previous M mark (\(a = 2\) and \(b = -4\))
A1ft: correct general solution following through their complementary function (must be their C.F. + \(2n - 4\)) Please award for a fully correct solution seen either here or being used in (b)
| Scheme | Marks | AO |
|---|---|---|
| \(A + B - 4 = 1\) \(0.5A - 3B - 2 = k\) leading to \(A = \ldots\) and \(B = \ldots\) | ddM1 | 3.4 |
| \((u_n =) \left(\dfrac{34 + 2k}{7}\right)(0.5)^n + \left(\dfrac{1 - 2k}{7}\right)(-3)^n + 2n - 4\) and setting their \(\dfrac{1 - 2k}{7} = 0\) | dddM1 | 3.1a |
| \(k = 0.5\) | A1 | 2.2a |
| (3) | ||
| (8 marks) |
Notes
ddM1: use correct initial conditions correctly to form simultaneous equations in their \(A\), \(B\) and \(k\) and attempt to solve for \(A\) and \(B\) – dependent on the previous two M marks. Award this mark for Either
- eliminating either \(A\) or \(B\) from the correct two equations leading to one of \(7A = 2k + 34\) or \(7B = 1 - 2k\) (accept any equivalent form)
Or
- eliminating both \(A\) and \(B\) from their equations using the initial conditions and obtaining expressions in terms of \(k\)
If correct \(A = \dfrac{34 + 2k}{7},\ B = \dfrac{1 - 2k}{7}\)
dddM1: setting coefficient (which must be a linear expression in \(k\)) of their \((-3)^n\) equal to zero – dependent on the previous three M marks (and at least one root of the auxiliary equation being negative)
A1: CAO (must be from correct working)