| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Standard +0.3 This is a standard second-order linear homogeneous recurrence relation with constant coefficients. The characteristic equation (r-3)² = 0 gives a repeated root, leading to the form u_n = (A + Bn)3^n. Finding A and B from initial conditions is routine FP2 material requiring only systematic application of the standard method with no novel insight. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Auxiliary equation: \(r^2 - 6r + 9 = 0 \Rightarrow r = \ldots\) | M1 | Forms and solves auxiliary equation |
| \((r-3)^2 = 0 \Rightarrow r = 3\) | A1 | Correct single root of 3 |
| General form is \(u_n = (A + Bn)3^n\) | M1 | Selects correct form for repeated root; if distinct roots found, allow \(u_n = A\alpha^n + B\beta^n\) |
| \(\begin{cases} 4 = 3(A+B) \\ 6 = 9(A+2B) \end{cases} \Rightarrow A = \ldots, B = \ldots\) | M1 | Uses \(u_1\) and \(u_2\) to form and solve simultaneous equations |
| \(u_n = \left(2 - \frac{2}{3}n\right)3^n\) or equivalently \(u_n = (6-2n)3^{n-1}\) | A1 | Correct closed form |
## Question 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Auxiliary equation: $r^2 - 6r + 9 = 0 \Rightarrow r = \ldots$ | M1 | Forms and solves auxiliary equation |
| $(r-3)^2 = 0 \Rightarrow r = 3$ | A1 | Correct single root of 3 |
| General form is $u_n = (A + Bn)3^n$ | M1 | Selects correct form for repeated root; if distinct roots found, allow $u_n = A\alpha^n + B\beta^n$ |
| $\begin{cases} 4 = 3(A+B) \\ 6 = 9(A+2B) \end{cases} \Rightarrow A = \ldots, B = \ldots$ | M1 | Uses $u_1$ and $u_2$ to form and solve simultaneous equations |
| $u_n = \left(2 - \frac{2}{3}n\right)3^n$ or equivalently $u_n = (6-2n)3^{n-1}$ | A1 | Correct closed form |
---\begin{enumerate}
\item Determine a closed form for the recurrence system
\end{enumerate}
$$\begin{gathered}
u _ { 1 } = 4 \quad u _ { 2 } = 6 \\
u _ { n + 2 } = 6 u _ { n + 1 } - 9 u _ { n } \quad ( n = 1,2,3 , \ldots )
\end{gathered}$$
\hfill \mbox{\textit{Edexcel FP2 2024 Q2 [5]}}