| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Second-Order Homogeneous Recurrence Relations |
| Difficulty | Standard +0.8 This is a Further Maths question on second-order recurrence relations requiring students to work backwards from the general solution form (A+Bn)(-3)^n to find the characteristic equation coefficients, then solve a system for specific initial conditions. While the technique is standard for FM students, the reverse-engineering aspect and the repeated root case (indicated by the (A+Bn) form) elevate it above routine drill exercises. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((m+3)^2 = 0\) | M1 | Correct auxiliary equation (may be implied by \(k_1=6\), \(k_2=9\) correct) from \(u_n=(A+Bn)(-3)^n\) |
| \(k_1 = 6\) and \(k_2 = 9\) | A1 | CAO; allow values stated implicitly i.e. \(u_{n+2}+6u_{n+1}+9u_n=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_0=1 \Rightarrow A(-3)^0=1\); \(u_1=1 \Rightarrow (A+B)(-3)=1\) | M1 | Uses \(u_0=u_1=1\) and attempts to find \(A\) and \(B\) |
| \(A=1\) and \(B=-\dfrac{4}{3}\) | A1 | CAO; allow stated implicitly e.g. \(u_n=\left(1-\dfrac{4}{3}n\right)(-3)^n\) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(m+3)^2 = 0$ | M1 | Correct auxiliary equation (may be implied by $k_1=6$, $k_2=9$ correct) from $u_n=(A+Bn)(-3)^n$ |
| $k_1 = 6$ and $k_2 = 9$ | A1 | CAO; allow values stated implicitly i.e. $u_{n+2}+6u_{n+1}+9u_n=0$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_0=1 \Rightarrow A(-3)^0=1$; $u_1=1 \Rightarrow (A+B)(-3)=1$ | M1 | Uses $u_0=u_1=1$ and attempts to find $A$ and $B$ |
| $A=1$ and $B=-\dfrac{4}{3}$ | A1 | CAO; allow stated implicitly e.g. $u_n=\left(1-\dfrac{4}{3}n\right)(-3)^n$ |
---2. The general solution of the second order recurrence relation
$$u _ { n + 2 } + k _ { 1 } u _ { n + 1 } + k _ { 2 } u _ { n } = 0 \quad n \geqslant 0$$
is given by
$$u _ { n } = ( A + B n ) ( - 3 ) ^ { n }$$
where $A$ and $B$ are arbitrary non-zero constants.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k _ { 1 }$ and the value of $k _ { 2 }$
Given that $u _ { 0 } = u _ { 1 } = 1$
\item find the value of $A$ and the value of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 2022 Q2 [4]}}