| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| 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 second-order linear recurrence with a polynomial non-homogeneous term. While it requires finding the complementary function (solving the characteristic equation), particular integral (trying u_n = an + b), and applying initial conditions, these are all standard FP2 techniques following a well-established algorithm with no novel insight required. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\lambda^2 = 2\lambda - \frac{4}{3} \Rightarrow 3\lambda^2 - 6\lambda + 4 = 0 \Rightarrow \lambda = \ldots\) | M1 | |
| \(\lambda = \frac{6 \pm \sqrt{36-48}}{6} = \frac{3 \pm i\sqrt{3}}{3} = \frac{2}{\sqrt{3}}e^{\pm i\frac{\pi}{6}}\) | A1 | |
| CF is \(u_n = A\left(\frac{3+i\sqrt{3}}{3}\right)^n + B\left(\frac{3-i\sqrt{3}}{3}\right)^n\) or \(\left(\frac{2\sqrt{3}}{3}\right)^n\left(P\cos\frac{\pi n}{6}+Q\sin\frac{\pi n}{6}\right)\) | A1ft | |
| \(u_n = kn+l \Rightarrow k(n+2)+l = 2k(n+1)+2l-\frac{4}{3}(kn+l)+n \Rightarrow \left(1-\frac{1}{3}k\right)n - \frac{1}{3}l = 0 \Rightarrow k=3, l=0\) | M1 | |
| \(u_n = A\left(\frac{3+i\sqrt{3}}{3}\right)^n + B\left(\frac{3-i\sqrt{3}}{3}\right)^n + 3n\) | A1 | |
| \(u_0=1 \Rightarrow A+B=1\); \(u_1=4 \Rightarrow A+B+(A-B)\frac{i\sqrt{3}}{3}=1 \Rightarrow (A-B)\frac{i\sqrt{3}}{3}=0 \Rightarrow A=B=\frac{1}{2}\) | M1 | Solves simultaneous equations for \(A,B\) or \(P,Q\) |
| \(u_n = \frac{1}{2}\left(\frac{3+i\sqrt{3}}{3}\right)^n + \frac{1}{2}\left(\frac{3-i\sqrt{3}}{3}\right)^n + 3n\) or \(u_n = \left(\frac{2\sqrt{3}}{3}\right)^n\cos\frac{\pi n}{6}+3n\) | A1 |
# Question 6:
| Working | Mark | Guidance |
|---------|------|----------|
| $\lambda^2 = 2\lambda - \frac{4}{3} \Rightarrow 3\lambda^2 - 6\lambda + 4 = 0 \Rightarrow \lambda = \ldots$ | M1 | |
| $\lambda = \frac{6 \pm \sqrt{36-48}}{6} = \frac{3 \pm i\sqrt{3}}{3} = \frac{2}{\sqrt{3}}e^{\pm i\frac{\pi}{6}}$ | A1 | |
| CF is $u_n = A\left(\frac{3+i\sqrt{3}}{3}\right)^n + B\left(\frac{3-i\sqrt{3}}{3}\right)^n$ or $\left(\frac{2\sqrt{3}}{3}\right)^n\left(P\cos\frac{\pi n}{6}+Q\sin\frac{\pi n}{6}\right)$ | A1ft | |
| $u_n = kn+l \Rightarrow k(n+2)+l = 2k(n+1)+2l-\frac{4}{3}(kn+l)+n \Rightarrow \left(1-\frac{1}{3}k\right)n - \frac{1}{3}l = 0 \Rightarrow k=3, l=0$ | M1 | |
| $u_n = A\left(\frac{3+i\sqrt{3}}{3}\right)^n + B\left(\frac{3-i\sqrt{3}}{3}\right)^n + 3n$ | A1 | |
| $u_0=1 \Rightarrow A+B=1$; $u_1=4 \Rightarrow A+B+(A-B)\frac{i\sqrt{3}}{3}=1 \Rightarrow (A-B)\frac{i\sqrt{3}}{3}=0 \Rightarrow A=B=\frac{1}{2}$ | M1 | Solves simultaneous equations for $A,B$ or $P,Q$ |
| $u_n = \frac{1}{2}\left(\frac{3+i\sqrt{3}}{3}\right)^n + \frac{1}{2}\left(\frac{3-i\sqrt{3}}{3}\right)^n + 3n$ or $u_n = \left(\frac{2\sqrt{3}}{3}\right)^n\cos\frac{\pi n}{6}+3n$ | A1 | |\begin{enumerate}
\item Determine a closed form for the recurrence relation
\end{enumerate}
$$\begin{aligned}
& u _ { 0 } = 1 \quad u _ { 1 } = 4 \\
& u _ { n + 2 } = 2 u _ { n + 1 } - \frac { 4 } { 3 } u _ { n } + n \quad n \geqslant 0
\end{aligned}$$
\hfill \mbox{\textit{Edexcel FP2 2023 Q6 [7]}}