| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Second-Order Homogeneous Recurrence Relations |
| Difficulty | Challenging +1.2 This is a structured second-order recurrence relation question with clear guidance through each part. Part (a) is straightforward translation of words to algebra, part (b) follows standard techniques for non-homogeneous recurrences (finding complementary function and particular integral), and part (c) requires identifying dominant terms as n→∞. While it requires multiple techniques and careful algebraic manipulation, the question scaffolds the solution and uses well-established methods from the Further Maths syllabus without requiring novel insight or particularly complex reasoning. |
| Spec | 8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_n - \frac{1}{2}(u_{n-1}+u_{n-2})=6 \Rightarrow 2u_n-(u_{n-1}+u_{n-2})=12\) | 1M1 | 2.1 — attempt to write given information as recurrence equation; allow sign errors and notation errors |
| \(2u_n - u_{n-1} - u_{n-2} = 12\) | 1A1 | 2.2a — CAO; equation provided in question so must see correct unsimplified form simplified with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Auxiliary equation \(2m^2 - m - 1 = 0 \Rightarrow m=1,\ m=-\frac{1}{2}\) | 1B1 | CAO for auxiliary equation and corresponding solutions (implied by correct complementary function) |
| \(u_n = A + B\left(-\frac{1}{2}\right)^n\) | 2B1 | Complementary function CAO; condone lack of '\(u_n=\)' |
| Particular solution: try \(u_n = \lambda n\); \(\therefore 2\lambda n - \lambda n + \lambda - \lambda n + 2\lambda = 12 \Rightarrow \lambda [=4]\) | 1M1 | Substitutes \(u_n = \lambda n\) into 2nd-order recurrence and solves for \(\lambda\); note \(2\lambda n - \lambda(n-1)-\lambda(n-2)=12\) o.e. can earn this mark |
| \(u_n = A + B\left(-\frac{1}{2}\right)^n + 4n\) | 1A1 | Correct general solution; condone lack of '\(u_n=\)' |
| \(u_1 = 2 \Rightarrow 2A - B = -4\) | 2M1 | Forms one equation in \(A\) and \(B\); general solution must be of form \(u_n = A+B\left(-\frac{1}{2}\right)^n + \mu n\) where \(\mu \neq 0\) |
| \(u_2 = 8 \Rightarrow 4A + B = 0\) | 3M1 | Forms second equation in \(A\) and \(B\); dependent on previous M mark |
| \(A = -\frac{2}{3},\ B = \frac{8}{3} \Rightarrow u_n = -\frac{2}{3} + \frac{8}{3}\left(-\frac{1}{2}\right)^n + 4n\) | 2A1 | 2.2a — particular solution CAO; do not condone lack of '\(u_n=\ldots\)' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_n \to 4n \quad (k=4)\) | 1M1 | 2.1 — obtains correct limit ft their particular solution (must be of form \(u_n = A+B\left(-\frac{1}{2}\right)^n + \mu n\), \(\mu \neq 0\)) |
| As \(n \to \infty\), \(\left(-\frac{1}{2}\right)^n \to 0\) and \(4n\) is considerably greater than \(-\frac{2}{3}\) | 1A1ft | 1.1b — provides correct reasoning including comments on both: (1) \(n\to\infty,\ \left(-\frac{1}{2}\right)^n\to 0\); (2) \(-\frac{2}{3}\) becomes negligible compared to \(4n\) as \(n\to\infty\). Condone '\(-\frac{2}{3}\) becomes insignificant' |
# Question 5 (Recurrence Relations):
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_n - \frac{1}{2}(u_{n-1}+u_{n-2})=6 \Rightarrow 2u_n-(u_{n-1}+u_{n-2})=12$ | 1M1 | 2.1 — attempt to write given information as recurrence equation; allow sign errors and notation errors |
| $2u_n - u_{n-1} - u_{n-2} = 12$ | 1A1 | 2.2a — CAO; equation provided in question so must see correct unsimplified form simplified with no errors |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Auxiliary equation $2m^2 - m - 1 = 0 \Rightarrow m=1,\ m=-\frac{1}{2}$ | 1B1 | CAO for auxiliary equation and corresponding solutions (implied by correct complementary function) |
| $u_n = A + B\left(-\frac{1}{2}\right)^n$ | 2B1 | Complementary function CAO; condone lack of '$u_n=$' |
| Particular solution: try $u_n = \lambda n$; $\therefore 2\lambda n - \lambda n + \lambda - \lambda n + 2\lambda = 12 \Rightarrow \lambda [=4]$ | 1M1 | Substitutes $u_n = \lambda n$ into 2nd-order recurrence and solves for $\lambda$; note $2\lambda n - \lambda(n-1)-\lambda(n-2)=12$ o.e. can earn this mark |
| $u_n = A + B\left(-\frac{1}{2}\right)^n + 4n$ | 1A1 | Correct general solution; condone lack of '$u_n=$' |
| $u_1 = 2 \Rightarrow 2A - B = -4$ | 2M1 | Forms one equation in $A$ and $B$; general solution must be of form $u_n = A+B\left(-\frac{1}{2}\right)^n + \mu n$ where $\mu \neq 0$ |
| $u_2 = 8 \Rightarrow 4A + B = 0$ | 3M1 | Forms second equation in $A$ and $B$; dependent on previous M mark |
| $A = -\frac{2}{3},\ B = \frac{8}{3} \Rightarrow u_n = -\frac{2}{3} + \frac{8}{3}\left(-\frac{1}{2}\right)^n + 4n$ | 2A1 | 2.2a — particular solution CAO; do not condone lack of '$u_n=\ldots$' |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_n \to 4n \quad (k=4)$ | 1M1 | 2.1 — obtains correct limit ft their particular solution (must be of form $u_n = A+B\left(-\frac{1}{2}\right)^n + \mu n$, $\mu \neq 0$) |
| As $n \to \infty$, $\left(-\frac{1}{2}\right)^n \to 0$ and $4n$ is considerably greater than $-\frac{2}{3}$ | 1A1ft | 1.1b — provides correct reasoning including comments on both: (1) $n\to\infty,\ \left(-\frac{1}{2}\right)^n\to 0$; (2) $-\frac{2}{3}$ becomes negligible compared to $4n$ as $n\to\infty$. Condone '$-\frac{2}{3}$ becomes insignificant' |
---5. An increasing sequence $\left\{ u _ { n } \right\}$ for $n \in \mathbb { N }$ is such that the difference between the $n$th term of $\left\{ u _ { n } \right\}$ and the mean of the previous two terms of $\left\{ u _ { n } \right\}$ is always 6
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 3$
$$2 u _ { n } - u _ { n - 1 } - u _ { n - 2 } = 12$$
Given that $u _ { 1 } = 2$ and $u _ { 2 } = 8$
\item find the solution of this second order recurrence relation to obtain an expression for $u _ { n }$ in terms of $n$.
\item Show that as $n \rightarrow \infty , u _ { n } \rightarrow k n$ where $k$ is a constant to be determined. You must give reasons for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 2019 Q5 [11]}}