| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Second-Order Homogeneous Recurrence Relations |
| Difficulty | Challenging +1.2 This is a Further Maths recurrence relation question requiring knowledge of complementary functions and particular integrals. Part (a) involves routine substitution into the auxiliary equation to find coefficients. Part (b) requires finding a particular integral (standard form for exponential RHS), then applying initial conditions to solve for constants. While this is Further Maths content and involves multiple steps, it follows a completely standard algorithmic procedure taught in FP2/FD2 with no novel problem-solving required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CF of \(u_n = A(2)_n + B(-1)_n \Rightarrow\) auxiliary equation is \((m-2)(m+1) = 0\) | M1 | Uses given complementary function to find auxiliary equation corresponding to second-order recurrence relation |
| \(m^2 - m - 2 = 0 \Rightarrow \alpha = -1,\ \beta = -2\) | A1 | cao for both \(\alpha\) and \(\beta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Particular solution: try \(u_n = \lambda(-3)_n\), \(u_{n+1} = \lambda(-3)_{n+1}\), \(u_{n+2} = \lambda(-3)_{n+2}\) | M1 | Substitute \(u_n = \lambda(-3)^n\) into second-order recurrence relation |
| \(9\lambda + 3\lambda - 2\lambda = 20 \Rightarrow \lambda = 2\) | M1 | Forms linear equation in \(\lambda\) only |
| \(u_n = A(2)^n + B(-1)^n + 2(-3)^n\) | A1 | Correct general solution |
| \(2u_0 = u_1 \Rightarrow 2A + 2B + 4 = 2A - B - 6\) | M1 | Use \(2u_0 = u_1\) to form equation in \(B\) (and possibly \(A\)) |
| \(u_4 = 164 \Rightarrow 16A + B + 162 = 164\) | M1 | Use \(u_4 = 164\) to set up second equation in \(A\) and \(B\) |
| \(A = \frac{1}{3},\ B = -\frac{10}{3} \Rightarrow u_n = \frac{1}{3}(2)^n - \frac{10}{3}(-1)^n + 2(-3)^n\) | A1 | cao |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CF of $u_n = A(2)_n + B(-1)_n \Rightarrow$ auxiliary equation is $(m-2)(m+1) = 0$ | M1 | Uses given complementary function to find auxiliary equation corresponding to second-order recurrence relation |
| $m^2 - m - 2 = 0 \Rightarrow \alpha = -1,\ \beta = -2$ | A1 | cao for both $\alpha$ and $\beta$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Particular solution: try $u_n = \lambda(-3)_n$, $u_{n+1} = \lambda(-3)_{n+1}$, $u_{n+2} = \lambda(-3)_{n+2}$ | M1 | Substitute $u_n = \lambda(-3)^n$ into second-order recurrence relation |
| $9\lambda + 3\lambda - 2\lambda = 20 \Rightarrow \lambda = 2$ | M1 | Forms linear equation in $\lambda$ only |
| $u_n = A(2)^n + B(-1)^n + 2(-3)^n$ | A1 | Correct general solution |
| $2u_0 = u_1 \Rightarrow 2A + 2B + 4 = 2A - B - 6$ | M1 | Use $2u_0 = u_1$ to form equation in $B$ (and possibly $A$) |
| $u_4 = 164 \Rightarrow 16A + B + 162 = 164$ | M1 | Use $u_4 = 164$ to set up second equation in $A$ and $B$ |
| $A = \frac{1}{3},\ B = -\frac{10}{3} \Rightarrow u_n = \frac{1}{3}(2)^n - \frac{10}{3}(-1)^n + 2(-3)^n$ | A1 | cao |
---\begin{enumerate}
\item The complementary function for the second order recurrence relation
\end{enumerate}
$$u _ { n + 2 } + \alpha u _ { n + 1 } + \beta u _ { n } = 20 ( - 3 ) ^ { n } \quad n \geqslant 0$$
is given by
$$u _ { n } = A ( 2 ) ^ { n } + B ( - 1 ) ^ { n }$$
where $A$ and $B$ are arbitrary non-zero constants.\\
(a) Find the value of $\alpha$ and the value of $\beta$.
Given that $2 u _ { 0 } = u _ { 1 }$ and $u _ { 4 } = 164$\\
(b) find the solution of this second order recurrence relation to obtain an expression for $u _ { n }$ in terms of $n$.\\
(6)\\
\hfill \mbox{\textit{Edexcel FD2 2020 Q4 [8]}}