| 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 | Applied recurrence modeling |
| Difficulty | Standard +0.3 This is a straightforward recurrence relation problem requiring students to form the relation un+1 = (3/4)un + k, solve it using standard techniques to get the general term, then substitute values. While it involves multiple steps, the methods are entirely standard for A-level Further Maths and require no novel insight—just systematic application of taught procedures. |
| Spec | 8.01a Recurrence relations: general sequences, closed form and recurrence8.01e Fibonacci: and related sequences (e.g. Lucas numbers)8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u_{n+1} = 0.75u_n + k\) | B1 | CAO |
| Auxiliary equation \(m - 0.75 = 0\), complementary function is \(A(0.75)^n\) | B1 | CAO |
| Trial solution \(u_n = \lambda\), so \(\lambda - 0.75\lambda = k\) | M1 | Substituting trial solution into recurrence relation of form \(u_{n+1} = \alpha u_n + k\) where \(\alpha = 0.75\) or \(\pm 0.25\), to find \(\lambda\) |
| General solution \(u_n = A(0.75)^n + 4k\) | A1 | CAO for general solution (may be implied by subsequent working) |
| \(u_1 = k\): \(0.75A + 4k = k \Rightarrow A = -3k\) | M1 | Using conditions in model with expression of form \(A(\alpha)^n + Bk\) where \(\alpha\) is \(0.75\) or \(\pm 0.25\), to calculate \(A\) (correct value is \(-4k\)) |
| \(u_n = 4k(1-(0.75)^n)\) | A1 | CAO for particular solution; accept equivalent e.g. \(u_n = 4k - 3k(0.75)^{n-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4k(1-0.75^8) = 1750 \Rightarrow k = \ldots\) | dM1 | Applying \(n=8\) to general solution of form \(A(\alpha)^n + Bk\) where \(\alpha\) is \(0.75\) or \(\pm 0.25\), set equal to 1750; dependent on both M marks in (a) |
| \(k = 486.1721068\ldots \Rightarrow k = 486.18\) | A1 | CAO — must be 486.18 |
## Question 7:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $u_{n+1} = 0.75u_n + k$ | B1 | CAO |
| Auxiliary equation $m - 0.75 = 0$, complementary function is $A(0.75)^n$ | B1 | CAO |
| Trial solution $u_n = \lambda$, so $\lambda - 0.75\lambda = k$ | M1 | Substituting trial solution into recurrence relation of form $u_{n+1} = \alpha u_n + k$ where $\alpha = 0.75$ or $\pm 0.25$, to find $\lambda$ |
| General solution $u_n = A(0.75)^n + 4k$ | A1 | CAO for general solution (may be implied by subsequent working) |
| $u_1 = k$: $0.75A + 4k = k \Rightarrow A = -3k$ | M1 | Using conditions in model with expression of form $A(\alpha)^n + Bk$ where $\alpha$ is $0.75$ or $\pm 0.25$, to calculate $A$ (correct value is $-4k$) |
| $u_n = 4k(1-(0.75)^n)$ | A1 | CAO for particular solution; accept equivalent e.g. $u_n = 4k - 3k(0.75)^{n-1}$ |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $4k(1-0.75^8) = 1750 \Rightarrow k = \ldots$ | dM1 | Applying $n=8$ to general solution of form $A(\alpha)^n + Bk$ where $\alpha$ is $0.75$ or $\pm 0.25$, set equal to 1750; dependent on both M marks in (a) |
| $k = 486.1721068\ldots \Rightarrow k = 486.18$ | A1 | CAO — must be 486.18 |7. Martina decides to open a bank account to help her to save for a holiday. Each month she puts $\pounds \mathrm { k }$ into the account and allows herself to spend one quarter of what was in the account at the end of the previous month.
Let $u _ { n }$ (where $\mathrm { n } \geqslant 1$ ) represent the amount in the account at the end of month n .\\
Martina has $\pounds \mathrm {~K}$ in the account at the end of the first month.
\begin{enumerate}[label=(\alph*)]
\item By setting up a first order recurrence relation for $u _ { n + 1 }$ in terms of $u _ { n }$, determine an expression for $u _ { n }$ in terms of n and k
At the end of the 8th month, Martina needs to have at least $\pounds 1750$ in the account to pay for her holiday.
\item Determine, to the nearest penny, the minimum amount of money that Martina should put into the account each month.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 2023 Q7 [8]}}