| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Modelling with Recurrence Relations |
| Difficulty | Standard +0.3 This is a standard recurrence relation question involving compound interest and regular payments. Part (a) requires forming a simple recurrence relation (routine). Part (b) is a straightforward proof by induction with algebraic manipulation that follows a standard template. Part (c) involves solving an inequality using the derived formula. While it requires multiple techniques (recurrence relations, induction, inequality solving), all steps follow well-practiced procedures with no novel insight required. This is slightly easier than average as it's a textbook application of standard FP2 techniques. |
| Spec | 4.01a Mathematical induction: construct proofs8.01a Recurrence relations: general sequences, closed form and recurrence8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A_{n+1} = 1.05A_n - F\) | B1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=1 \Rightarrow A_1 = (10\,000 - 20F)1.05^{1-1} + 20F = 10\,000 - 20F + 20F = 10\,000\). So true for \(n=1\) | B1 | Demonstrates that the result is true for \(n=1\) |
| Assume true for \(n = k\) so that \(A_k = (10\,000 - 20F)1.05^{k-1} + 20F\) | M1 | Makes a statement that assumes the result is true for some value of \(n\) |
| \(A_{k+1} = 1.05\bigl((10\,000 - 20F)1.05^{k-1} + 20F\bigr) - F\) | A1ft | Correct expression for \(A_{k+1}\) |
| \(A_{k+1} = (10\,000 - 20F)1.05^k + 21F - F = (10\,000 - 20F)1.05^k + 20F\) | A1 | Reaches the correct statement for \(k+1\) in the required form |
| So the result holds for \(n = k+1\) and so \(A_n = (10\,000 - 20F)1.05^{n-1} + 20F\) is true for all \(n \geq 1\) | A1 | Completes the inductive argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((10\,000 - 20F)1.05^{16-1} + 20F \leq 0\) | M1 | Identifies correct strategy using \(n=16\) to obtain an equation in \(F\) |
| \(10\,000 \times 1.05^{15} \leq 20F(1.05^{15} - 1) \Rightarrow F \geq \ldots\) | M1 | Proceeds to obtain a value for the minimum value of \(F\) |
| \(F \geq \dfrac{10\,000 \times 1.05^{15}}{20(1.05^{15} - 1)}\) | A1 | Correct numerical expression for \(F\) |
| Smallest value of \(F\) is £963.43 or £964 | A1 | Correct answer (allow to nearest penny or nearest £) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A_{n+1} = 1.05A_n - F$ | B1 | Correct equation |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1 \Rightarrow A_1 = (10\,000 - 20F)1.05^{1-1} + 20F = 10\,000 - 20F + 20F = 10\,000$. So true for $n=1$ | B1 | Demonstrates that the result is true for $n=1$ |
| Assume true for $n = k$ so that $A_k = (10\,000 - 20F)1.05^{k-1} + 20F$ | M1 | Makes a statement that assumes the result is true for some value of $n$ |
| $A_{k+1} = 1.05\bigl((10\,000 - 20F)1.05^{k-1} + 20F\bigr) - F$ | A1ft | Correct expression for $A_{k+1}$ |
| $A_{k+1} = (10\,000 - 20F)1.05^k + 21F - F = (10\,000 - 20F)1.05^k + 20F$ | A1 | Reaches the correct statement for $k+1$ in the required form |
| So the result holds for $n = k+1$ and so $A_n = (10\,000 - 20F)1.05^{n-1} + 20F$ is true for all $n \geq 1$ | A1 | Completes the inductive argument |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(10\,000 - 20F)1.05^{16-1} + 20F \leq 0$ | M1 | Identifies correct strategy using $n=16$ to obtain an equation in $F$ |
| $10\,000 \times 1.05^{15} \leq 20F(1.05^{15} - 1) \Rightarrow F \geq \ldots$ | M1 | Proceeds to obtain a value for the minimum value of $F$ |
| $F \geq \dfrac{10\,000 \times 1.05^{15}}{20(1.05^{15} - 1)}$ | A1 | Correct numerical expression for $F$ |
| Smallest value of $F$ is £963.43 or £964 | A1 | Correct answer (allow to nearest penny or nearest £) |
---\begin{enumerate}
\item Sam borrows $\pounds 10000$ from a bank to pay for an extension to his house.
\end{enumerate}
The bank charges $5 \%$ annual interest on the portion of the loan yet to be repaid. Immediately after the interest has been added at the end of each year and before the start of the next year, Sam pays the bank a fixed amount, $\pounds F$.
Given that $\pounds A _ { n }$ (where $A _ { n } \geqslant 0$ ) is the amount owed at the start of year $n$,\\
(a) write down an expression for $A _ { n + 1 }$ in terms of $A _ { n }$ and $F$,\\
(b) prove, by induction that, for $n \geqslant 1$
$$A _ { n } = ( 10000 - 20 F ) 1.05 ^ { n - 1 } + 20 F$$
(c) Find the smallest value of $F$ for which Sam can repay all of the loan by the start of year 16 .
\hfill \mbox{\textit{Edexcel FP2 AS 2020 Q4 [10]}}