# Question 4:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Auxiliary equation $m + 3 = 0$; complementary function is $A(-3)^n$ | B1 | cao; condone $A(-3)^{n-1}$; condone $A(-3)^n$ in (a) except for final A mark |
| Particular solution: try $u_n = an + b$, substitute into recurrence relation | M1 | Correct form substituted into recurrence relation |
| $a(n+1) + b + 3(an+b) = n + k$; comparing terms gives $a + 3a = 1$ and $a + b + 3b = k$ | dM1 | Compare coefficients; set up both equations in $a, b, k$; dependent on previous M |
| $a = \dfrac{1}{4},\quad 4b + \dfrac{1}{4} = k \Rightarrow b = \dfrac{4k-1}{16}$ | ddM1 | Solve for $a$ and $b$ (with $b$ in terms of $k$); dependent on both previous M marks |
| $u_n = A(-3)^n + \dfrac{1}{4}n + \dfrac{4k-1}{16}$ | A1 | Correct general solution in terms of $k$; ignore labelling of LHS |
| $u_0 = 1 \Rightarrow A + \dfrac{4k-1}{16} = 1$ | dddM1 | Use initial condition correctly to form equation in $A$ and $k$; dependent on all three previous M marks |
| $u_n = \left(\dfrac{17-4k}{16}\right)(-3)^n + \dfrac{1}{4}n + \dfrac{4k-1}{16}$ | A1 | Correct particular solution in terms of $k$; must have correct LHS |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Setting $\dfrac{17-4k}{16} = 0$ and solving for $k$ | M1 | |
| $k = \dfrac{17}{4} \Rightarrow u_{100} = \dfrac{1}{4}(100) + 1$ | dM1 | Dependent on previous M mark |
| $u_{100} = 26$ | A1 | cao |

# Question (b):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Setting the coefficient of the exponential term equal to zero and solving for $k$ – 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) | M1 | Setting coefficient of exponential term to zero and solving for $k$ |
| Substituting their value of $k$ to obtain an expression for $u_n$ which is linear and substituting $n = 100$ | dM1 | Dependent on previous M mark |
| cao $(26)$ – from correct working including a correct expression for $u_n$ in (a) | A1 | |

**Additional Guidance:**

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Those who rewrite $u_{n+1} + 3u_n = n+k$ as $u_n + 3u_{n-1} = (n-1)+k$ can score full marks. CF is $A(-3)^n$, PS of form $an+b$ leading to $4an + 4b - 3a - n + 1 - k = 0$ and so $a = \frac{1}{4}, b = \frac{4k-1}{16}$ | | Full marks available via this route |
| Rewriting $u_{n+1} + 3u_n = n+k$ as $u_n + 3u_{n-1} = n+k$ is **incorrect** | | Can score all B and M marks in both parts only |
| CF of form $A(-3)^{n-1}$, PS of form $a(n-1)+b$ 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 $b = \frac{4k+7}{16}$ giving $(u_{n+1} =) A(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}$ | | Alternative valid approach |
| Using $u_0 = 1 \Rightarrow u_1 = k-3$ gives $A = \frac{153-36k}{16}$, so $u_{n+1} = \left(\frac{153-36k}{16}\right)(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}$ which re-written in terms of $u_n$ gives form as in main mark scheme | | |
| Any CF of form $A(-3)^{n \pm k_1}$ and any PS of form $a(n \pm k_2)+b$ where $k_1, k_2$ are constants will work | | Award M marks for correct methods; first A mark in (a) for correct general solution ignoring labelling of LHS; second A mark in (a) for fully correct expression with correct LHS |