| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | March |
| Marks | 15 |
| Topic | Sequences and Series |
| Type | Second-Order Homogeneous Recurrence Relations |
| Difficulty | Challenging +1.8 This is a Further Maths recurrence relation question requiring characteristic equation solution, asymptotic analysis, and applied interpretation. While the mechanics are standard (solving auxiliary equation, finding particular solution), parts (i)(b) requires recognizing dominant term behavior, part (ii) involves logarithmic calculation, and part (iii) demands conceptual understanding of rounding effects. The multi-step nature, combination of pure and applied reasoning, and Further Maths context place this well above average difficulty. |
| Spec | 8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states8.01g Second-order recurrence: solve with distinct, repeated, or complex roots8.01h Modelling with recurrence: birth/death rates, INT function8.01i Modelling with second-order: recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Characteristic Equation is \(\lambda^2 - 1.3\lambda - 0.3 = 0\) \(\Rightarrow \lambda = 1.5, -0.2\) | M1, A1 | BC |
| \(\Rightarrow\) General Solution is \(X_n = A \times 1.5^n + B \times (-0.2)^n\) | B1 | FT their \(\lambda\)s |
| Use of \(X_0 = 12\) and \(X_1 = 1\) to obtain equations in \(A, B\): i.e. \(12 = A + B\) & \(1 = 1.5A - 0.2B\) | M1 | Eqns. solved simultaneously (e.g. BC) |
| Solving \(\Rightarrow A = 2, B = 10\) | M1 | |
| Solution is \(X_n = 2 \times 1.5^n + 10 \times (-0.2)^n\) | A1 [6] | cao brackets required |
| Answer | Marks | Guidance |
|---|---|---|
| For large \(n\), \((-0.2)^n \to 0\) so \(X_n \to 2 \times 1.5^n\) which is of the form \(ar^n\), hence a GP | B1, E1 [2] | OR \(3 \times 1.5^{n-1}\), of the form \(ar^{n-1}\), hence a GP [MUST have some explanation that this is a GP] |
| Answer | Marks | Guidance |
|---|---|---|
| Tabulating the given sequence (or using calculator equation solver) \(X_{32} \approx 862 \, 880 < 1 \, 000 \, 000\) and \(X_{33} \approx 1 \, 294 \, 320 > 1 \, 000 \, 000\) so \(n = 33\) | M1, A1 [2] | BC or manual calculation; Properly justified |
| Answer | Marks | Guidance |
|---|---|---|
| \(X_n = 2 \times 1.5^n > 1 \, 000 \, 000\) \(\Rightarrow n > \frac{\log 500000}{\log 1.5} = 32.36...\) so \(n = 33\) | M1, A1 [2] | Attempt at solving (logs not essential); Properly justified |
| Answer | Marks |
|---|---|
| \(X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n]\) | B1 [1] |
| Answer | Marks |
|---|---|
| \(X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n + 1]\) or \(X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n] + 1\) | B1 [1] |
| Answer | Marks |
|---|---|
| If \(1.3X_{n+1} + 0.3X_n\) were exactly an integer value at any stage, then the next \(X_n\) would be too large by 1 | B1 [1] |
## (i) (a)
| Characteristic Equation is $\lambda^2 - 1.3\lambda - 0.3 = 0$ $\Rightarrow \lambda = 1.5, -0.2$ | M1, A1 | BC |
| $\Rightarrow$ General Solution is $X_n = A \times 1.5^n + B \times (-0.2)^n$ | B1 | FT their $\lambda$s |
| Use of $X_0 = 12$ and $X_1 = 1$ to obtain equations in $A, B$: i.e. $12 = A + B$ & $1 = 1.5A - 0.2B$ | M1 | Eqns. solved simultaneously (e.g. BC) |
| Solving $\Rightarrow A = 2, B = 10$ | M1 | |
| Solution is $X_n = 2 \times 1.5^n + 10 \times (-0.2)^n$ | A1 [6] | cao brackets required |
## (i) (b)
| For large $n$, $(-0.2)^n \to 0$ so $X_n \to 2 \times 1.5^n$ which is of the form $ar^n$, hence a GP | B1, E1 [2] | OR $3 \times 1.5^{n-1}$, of the form $ar^{n-1}$, hence a GP [MUST have some explanation that this is a GP] |
## (ii) (a)
| Tabulating the given sequence (or using calculator equation solver) $X_{32} \approx 862 \, 880 < 1 \, 000 \, 000$ and $X_{33} \approx 1 \, 294 \, 320 > 1 \, 000 \, 000$ so $n = 33$ | M1, A1 [2] | BC or manual calculation; Properly justified |
## (ii) (b)
| $X_n = 2 \times 1.5^n > 1 \, 000 \, 000$ $\Rightarrow n > \frac{\log 500000}{\log 1.5} = 32.36...$ so $n = 33$ | M1, A1 [2] | Attempt at solving (logs not essential); Properly justified |
## (iii) (a)
| $X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n]$ | B1 [1] | |
## (iii) (b)
| $X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n + 1]$ or $X_{n+2} = \text{INT}[1.3X_{n+1} + 0.3X_n] + 1$ | B1 [1] | |
## (iii) (c)
| If $1.3X_{n+1} + 0.3X_n$ were exactly an integer value at any stage, then the next $X_n$ would be too large by 1 | B1 [1] | |
---5
\begin{enumerate}[label=(\roman*)]
\item (a) Solve the recurrence relation
$$X _ { n + 2 } = 1.3 X _ { n + 1 } + 0.3 X _ { n } \text { for } n \geqslant 0$$
given that $X _ { 0 } = 12$ and $X _ { 1 } = 1$.\\
(b) Show that the sequence $\left\{ X _ { n } \right\}$ approaches a geometric sequence as $n$ increases.
The recurrence relation in part (i) models the projected annual profit for an investment company, so that $X _ { n }$ represents the profit (in $\pounds$ ) at the end of year $n$.
\item (a) Determine the number of years taken for the projected profit to exceed one million pounds.\\
(b) Compare your answer to part (ii)(a) with the corresponding figure given by the geometric sequence of part (i)(b).
\item (a) In a modified model, any non-integer values obtained are rounded down to the nearest integer at each step of the process. Write down the recurrence relation for this model.\\
(b) Write down the recurrence relation for the model in which any non-integer values obtained are rounded up to the nearest integer at each step of the process.\\
(c) Describe a situation that might arise in the implementation of part (iii)(b) that would result in an incorrect value for the next $X _ { n }$ in the process.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q5 [15]}}