| Exam Board | Edexcel |
|---|---|
| Module | FD2 (Further Decision 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| 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 structured second-order recurrence relation problem where the substitution is explicitly given, removing the need for insight. Students follow guided steps: substitute to eliminate the non-homogeneous term, solve the resulting homogeneous recurrence using standard auxiliary equation methods, then back-substitute. Part (b) requires analyzing the limit behavior from the closed form. While this involves multiple techniques (substitution, auxiliary equations, limit analysis), the scaffolding makes it more routine than problems requiring independent problem-solving or proof construction. |
| Spec | 8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states8.01f First-order recurrence: solve using auxiliary equation and complementary function8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substituting \(u_n = v_n + 2n\) to get: \(v_{n+2} + 2(n+2) = \frac{1}{2}(v_{n+1} + 2(n+1) + v_n + 2n) + 3\) | M1 | 2.1 — Substituting \(u_n = v_n + 2n\) to obtain a second-order recurrence relation in \(v_n\) only |
| \(2v_{n+2} - v_{n+1} - v_n = 0\) | A1 | 1.1b — CAO Correct homogeneous second-order recurrence relation |
| \(2m^2 - m - 1 = 0 \Rightarrow (2m+1)(m-1) = 0 \therefore m = -0.5\) and \(1\) | dM1 | 1.1b — Solving three-term auxiliary equation (dependent on previous M mark) |
| \(v_n = A + B\left(-\frac{1}{2}\right)^n\) | A1ft | 1.1b — Correct general solution for \(v_n\) following their roots |
| \(u_0 = 15 \Rightarrow v_0 = 15\); \(u_1 = 20 \Rightarrow v_1 = 18\) | B1 | 1.1b — Converting first two given values for \(u\) to \(v\), or stating corresponding general solution for \(u_n\) |
| \(A + B = 15\) and \(A - \frac{1}{2}B = 18\), solve for \(A\) and \(B\) | M1 | 1.1b — Setting up two equations and solving for \(A\) and \(B\) (if correct \(A=17\), \(B=-2\)), dependent on general solution of form \(A + B(\alpha)^n\) where \(\alpha\) is \(-\frac{1}{2}\) or \(-2\) |
| \(u_n = 17 - 2\left(-\frac{1}{2}\right)^n + 2n\) | A1 | 2.2a — CAO; accept equivalent forms e.g. \(u_n = 17 + \left(-\frac{1}{2}\right)^{n-1} + 2n\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2u_{n+2} - u_{n+1} - u_n = 6\) | M0, A0 | Transformation not used |
| \(2u_{n+2} - u_{n+1} - u_n = 0\); auxiliary: \(2m^2 - m - 1 = 0 \Rightarrow m = -0.5, 1\) | M1 | Setting up and solving three-term auxiliary equation |
| C.F. \(u_n = A + B\left(-\frac{1}{2}\right)^n\) | A1ft | Correct complementary function following their roots |
| Trial solution \(\lambda n\): \(\lambda(n+2) - \frac{1}{2}\lambda(n+1) - \frac{1}{2}\lambda n = 3 \Rightarrow \lambda = 2\), P.S. \(2n\) | B1 | Finding correct particular solution and stating general solution \(u_n = A + B\left(-\frac{1}{2}\right)^n + 2n\) |
| \(A + B = 15\), \(A - \frac{1}{2}B = 18\), solve for \(A\) and \(B\) | dM1 | Setting up two equations; dependent on general solution of form \(A + B(\alpha)^n + Cn\) |
| \(u_n = 17 - 2\left(-\frac{1}{2}\right)^n + 2n\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(n \to \infty\), the terms of \(u_n\) are (approximately) given by the linear expression \(17 + 2n\) | B1 | 2.4 — Explanation that as \(n\) becomes large the terms are approximately in arithmetic progression (or equivalent); dependent on correct expression for \(u_n\) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substituting $u_n = v_n + 2n$ to get: $v_{n+2} + 2(n+2) = \frac{1}{2}(v_{n+1} + 2(n+1) + v_n + 2n) + 3$ | M1 | 2.1 — Substituting $u_n = v_n + 2n$ to obtain a second-order recurrence relation in $v_n$ only |
| $2v_{n+2} - v_{n+1} - v_n = 0$ | A1 | 1.1b — CAO Correct homogeneous second-order recurrence relation |
| $2m^2 - m - 1 = 0 \Rightarrow (2m+1)(m-1) = 0 \therefore m = -0.5$ and $1$ | dM1 | 1.1b — Solving three-term auxiliary equation (dependent on previous M mark) |
| $v_n = A + B\left(-\frac{1}{2}\right)^n$ | A1ft | 1.1b — Correct general solution for $v_n$ following their roots |
| $u_0 = 15 \Rightarrow v_0 = 15$; $u_1 = 20 \Rightarrow v_1 = 18$ | B1 | 1.1b — Converting first two given values for $u$ to $v$, or stating corresponding general solution for $u_n$ |
| $A + B = 15$ and $A - \frac{1}{2}B = 18$, solve for $A$ and $B$ | M1 | 1.1b — Setting up two equations and solving for $A$ and $B$ (if correct $A=17$, $B=-2$), dependent on general solution of form $A + B(\alpha)^n$ where $\alpha$ is $-\frac{1}{2}$ or $-2$ |
| $u_n = 17 - 2\left(-\frac{1}{2}\right)^n + 2n$ | A1 | 2.2a — CAO; accept equivalent forms e.g. $u_n = 17 + \left(-\frac{1}{2}\right)^{n-1} + 2n$ |
**Special Case (no transformation used): Max 6/8**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2u_{n+2} - u_{n+1} - u_n = 6$ | M0, A0 | Transformation not used |
| $2u_{n+2} - u_{n+1} - u_n = 0$; auxiliary: $2m^2 - m - 1 = 0 \Rightarrow m = -0.5, 1$ | M1 | Setting up and solving three-term auxiliary equation |
| C.F. $u_n = A + B\left(-\frac{1}{2}\right)^n$ | A1ft | Correct complementary function following their roots |
| Trial solution $\lambda n$: $\lambda(n+2) - \frac{1}{2}\lambda(n+1) - \frac{1}{2}\lambda n = 3 \Rightarrow \lambda = 2$, P.S. $2n$ | B1 | Finding correct particular solution and stating general solution $u_n = A + B\left(-\frac{1}{2}\right)^n + 2n$ |
| $A + B = 15$, $A - \frac{1}{2}B = 18$, solve for $A$ and $B$ | dM1 | Setting up two equations; dependent on general solution of form $A + B(\alpha)^n + Cn$ |
| $u_n = 17 - 2\left(-\frac{1}{2}\right)^n + 2n$ | A1 | CAO |
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## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $n \to \infty$, the terms of $u_n$ are (approximately) given by the linear expression $17 + 2n$ | B1 | 2.4 — Explanation that as $n$ becomes large the terms are approximately in arithmetic progression (or equivalent); dependent on correct expression for $u_n$ |
---5. A sequence $\left\{ u _ { n } \right\}$, where $\mathrm { n } \geqslant 0$, satisfies the second order recurrence relation
$$u _ { n + 2 } = \frac { 1 } { 2 } \left( u _ { n + 1 } + u _ { n } \right) + 3 \text { where } u _ { 0 } = 15 \quad u _ { 1 } = 20$$
\begin{enumerate}[label=(\alph*)]
\item By considering the sequence $\left\{ v _ { n } \right\}$, where $u _ { n } = v _ { n } + 2 n$ for $\mathrm { n } \geqslant 0$, determine an expression for $u _ { n }$ as a function of n .
\item Describe the long-term behaviour of $u _ { n }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 2023 Q5 [8]}}