| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.2 This is a standard Further Maths recurrence relation problem requiring substitution to derive a second-order relation, solving the homogeneous equation using characteristic equation methods, finding a particular solution, and applying initial conditions. While it involves multiple steps and is from Further Maths (inherently harder), the techniques are routine and well-practiced for FP students—no novel insight required, just systematic application of standard methods. |
| Spec | 8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x_{n+2} = 2y_{n+1}+3 \Rightarrow x_{n+2} = 2(-4x_n+3x_{n+1})+3\) leading to \(x_{n+2}-6x_{n+1}+8x_n=3\) | B1 | Correct reasoning to derive given result – sufficient working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Auxiliary equation \(m^2-6m+8=0 \Rightarrow m=2, m=4\) | B1 | cao for auxiliary equation and corresponding solutions |
| \(x_n = A(2)^n + B(4)^n\) | B1 | cao for the complementary function |
| Particular solution try \(x_n = \lambda\); \(\therefore \lambda - 6\lambda + 8\lambda = 3 \Rightarrow \lambda=1\) | M1 | Substitute \(x_n=\lambda\) into second order recurrence relation and solve for \(\lambda\) |
| \(x_n = A(2)^n + B(4)^n + 1\) | A1 | Correct general solution |
| \(x_1=1 \Rightarrow 2A+4B=0\) | M1 | Forms one equation in \(A\) and \(B\) using \(x_1=1\) |
| \(y_1=a \Rightarrow x_2=2a+3\) | B1 | Uses original recurrence relation for \(x_{n+1}\) to derive the expression \(2a+3\) |
| \(4A+16B+1=2a+3\) | M1 | Setting up a second equation in \(A\) and \(B\) |
| \(A=-\dfrac{(a+1)}{2},\ B=\dfrac{(a+1)}{4} \Rightarrow x_n=(a+1)(4)^{n-1}-(a+1)(2)^{n-1}+1\) | A1 | cao (oe e.g., \(x_n=0.25(a+1)(4)^n - 0.5(a+1)(2)^n+1\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| As \(x_7=28225 \Rightarrow (a+1)(4)^6-(a+1)(2)^6+1=28225\) leading to \(a=\ldots\) | M1 | Using \(x_7=28225\) to form a linear equation in \(a\) and attempt to solve for \(a\) |
| \(a=6\) | A1 | cao for \(a\) |
# Question 4:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x_{n+2} = 2y_{n+1}+3 \Rightarrow x_{n+2} = 2(-4x_n+3x_{n+1})+3$ leading to $x_{n+2}-6x_{n+1}+8x_n=3$ | B1 | Correct reasoning to derive given result – sufficient working must be shown |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Auxiliary equation $m^2-6m+8=0 \Rightarrow m=2, m=4$ | B1 | cao for auxiliary equation and corresponding solutions |
| $x_n = A(2)^n + B(4)^n$ | B1 | cao for the complementary function |
| Particular solution try $x_n = \lambda$; $\therefore \lambda - 6\lambda + 8\lambda = 3 \Rightarrow \lambda=1$ | M1 | Substitute $x_n=\lambda$ into second order recurrence relation and solve for $\lambda$ |
| $x_n = A(2)^n + B(4)^n + 1$ | A1 | Correct general solution |
| $x_1=1 \Rightarrow 2A+4B=0$ | M1 | Forms one equation in $A$ and $B$ using $x_1=1$ |
| $y_1=a \Rightarrow x_2=2a+3$ | B1 | Uses original recurrence relation for $x_{n+1}$ to derive the expression $2a+3$ |
| $4A+16B+1=2a+3$ | M1 | Setting up a second equation in $A$ and $B$ |
| $A=-\dfrac{(a+1)}{2},\ B=\dfrac{(a+1)}{4} \Rightarrow x_n=(a+1)(4)^{n-1}-(a+1)(2)^{n-1}+1$ | A1 | cao (oe e.g., $x_n=0.25(a+1)(4)^n - 0.5(a+1)(2)^n+1$) |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| As $x_7=28225 \Rightarrow (a+1)(4)^6-(a+1)(2)^6+1=28225$ leading to $a=\ldots$ | M1 | Using $x_7=28225$ to form a linear equation in $a$ and attempt to solve for $a$ |
| $a=6$ | A1 | cao for $a$ |
---\begin{enumerate}
\item Sequences $\left\{ x _ { n } \right\}$ and $\left\{ y _ { n } \right\}$ for $n \in \mathbb { N }$, are defined by
\end{enumerate}
$$\begin{gathered}
x _ { n + 1 } = 2 y _ { n } + 3 \quad \text { and } \quad y _ { n + 1 } = 3 x _ { n + 1 } - 4 x _ { n } \\
x _ { 1 } = 1 \quad \text { and } \quad y _ { 1 } = a
\end{gathered}$$
where $a$ is a constant.\\
(a) Show that $x _ { n + 2 } - 6 x _ { n + 1 } + 8 x _ { n } = 3$\\
(b) Solve the second-order recurrence relation given in (a) to obtain an expression for $x _ { n }$ in terms of $a$ and $n$.
Given that $x _ { 7 } = 28225$\\
(c) find the value of $a$.\\
\hfill \mbox{\textit{Edexcel FD2 2021 Q4 [11]}}