| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation asymptotic behaviour |
| Difficulty | Challenging +1.8 This is a non-homogeneous linear recurrence relation requiring the complementary function plus particular integral method, which is advanced A-level content. Part (a) demands systematic algebraic manipulation to find the general solution (trying particular solutions of form an²+bn+c), while part (b) is straightforward substitution once (a) is solved. The multi-step nature, algebraic complexity, and requirement for a structured approach to non-homogeneous recurrences places this well above average difficulty. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Auxiliary equation \(m - \frac{3}{2} = 0 \Rightarrow\) complementary function is \(A(1.5)^n\) | B1 | cao (or equivalent e.g. \(A(1.5)^{n-1}\)) |
| Particular solution try \(u_n = \alpha n^2 + \beta n + \gamma\) and substitute into recurrence relation | M1 | Correct form for particular solution e.g. \(\alpha n^2 + \beta n + \gamma\), \(\alpha(n-1)^2 + \beta(n-1) + \gamma\) etc. (three term quadratic in \(n\)) together with valid substitution into recurrence relation (not substituting \(u_n\) on both sides) |
| \(2\alpha n^2 + (4\alpha + 2\beta)n + (2\alpha + 2\beta + 2\gamma) = (3\alpha - 4)n^2 + 3\beta n + (3\gamma - 8)\) giving: \(2\alpha = 3\alpha - 4\), \(\quad 4\alpha + 2\beta = 3\beta\), \(\quad 2\alpha + 2\beta + 2\gamma = 3\gamma - 8\) | dM1 | Compares coefficients and sets up all three equations in \(\alpha, \beta, \gamma\) — dependent on previous M mark |
| \(\alpha = 4,\ \beta = 16,\ \gamma = 48\) | A1 | cao for the values of \(\alpha, \beta, \gamma\) |
| \((u_n =)\ A(1.5)^n + 4n^2 + 16n + 48\) | — | — |
| \(u_0 = k \Rightarrow A + 48 = k\) | ddM1 | Use correct initial condition to form linear equation in \(A\) and \(k\) — dependent on the two previous M marks |
| \((u_n =)\ (k-48)(1.5)^n + 4n^2 + 16n + 48\) | A1 | Correct particular solution in terms of \(k\) — need not see \(u_n =\) but if seen must be correct (and therefore \(u_{n+1} =\) is A0) |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((k-48)(1.5)^{10} + 4(10)^2 + 16(10) + 48 > 5000\) | M1 | Dependent on all M marks in (a) — substituting \(n = 10\) and setting the particular solution \(> 5000\) (or equal to) |
| \(k > 124.163\ldots \Rightarrow k = 125\) | A1 | cao (125) from correct working — must have had correct expression for \(u_n\) in (a) (dependent on at least first 5 marks in (a)) |
| (2) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Auxiliary equation $m - \frac{3}{2} = 0 \Rightarrow$ complementary function is $A(1.5)^n$ | B1 | cao (or equivalent e.g. $A(1.5)^{n-1}$) |
| Particular solution try $u_n = \alpha n^2 + \beta n + \gamma$ and substitute into recurrence relation | M1 | Correct form for particular solution e.g. $\alpha n^2 + \beta n + \gamma$, $\alpha(n-1)^2 + \beta(n-1) + \gamma$ etc. (three term quadratic in $n$) **together** with valid substitution into recurrence relation (not substituting $u_n$ on both sides) |
| $2\alpha n^2 + (4\alpha + 2\beta)n + (2\alpha + 2\beta + 2\gamma) = (3\alpha - 4)n^2 + 3\beta n + (3\gamma - 8)$ giving: $2\alpha = 3\alpha - 4$, $\quad 4\alpha + 2\beta = 3\beta$, $\quad 2\alpha + 2\beta + 2\gamma = 3\gamma - 8$ | dM1 | Compares coefficients and sets up all three equations in $\alpha, \beta, \gamma$ — dependent on previous M mark |
| $\alpha = 4,\ \beta = 16,\ \gamma = 48$ | A1 | cao for the values of $\alpha, \beta, \gamma$ |
| $(u_n =)\ A(1.5)^n + 4n^2 + 16n + 48$ | — | — |
| $u_0 = k \Rightarrow A + 48 = k$ | ddM1 | Use correct initial condition to form linear equation in $A$ and $k$ — dependent on the two previous M marks |
| $(u_n =)\ (k-48)(1.5)^n + 4n^2 + 16n + 48$ | A1 | Correct particular solution in terms of $k$ — need not see $u_n =$ but if seen must be correct (and therefore $u_{n+1} =$ is **A0**) |
| | **(6)** | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(k-48)(1.5)^{10} + 4(10)^2 + 16(10) + 48 > 5000$ | M1 | Dependent on all M marks in (a) — substituting $n = 10$ and setting the particular solution $> 5000$ (or equal to) |
| $k > 124.163\ldots \Rightarrow k = 125$ | A1 | cao (125) from correct working — must have had correct expression for $u_n$ in (a) (dependent on at least first 5 marks in (a)) |
| | **(2)** | |4. A sequence $\left\{ u _ { n } \right\}$, where $n \geqslant 0$, satisfies the recurrence relation
$$u _ { n + 1 } = \frac { 3 } { 2 } u _ { n } - 2 n ^ { 2 } - 4 \quad u _ { 0 } = k$$
where $k$ is an integer.
\begin{enumerate}[label=(\alph*)]
\item Determine an expression for $u _ { n }$ in terms of $n$ and $k$.\\
(6)
Given that $u _ { 10 } > 5000$
\item determine the minimum possible value of $k$.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 AS 2023 Q4 [8]}}