| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| 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 FP2 recurrence relation question requiring: (a) writing down a simple recurrence relation from a word problem, (b) verifying a given closed form solution by substitution and checking initial conditions, and (c) solving an exponential inequality using logarithms. All steps are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V_{n+2} = V_{n+1} + kV_n\) | B1 | Correct expression for model using information given |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda^2 - \lambda - 0.24 = 0 \Rightarrow \lambda = ...(1.2, -0.2)\) | M1 | Forms and solves auxiliary equation with \(k=0.24\) |
| \(V_n = a(1.2)^n + b(-0.2)^n\) | A1 | Correct closed form; not a follow-through mark |
| \(65 = a(1.2)^1 + b(-0.2)^1\) and \(71 = a(1.2)^2 + b(-0.2)^2\) | B1ft | Applies initial conditions \(V_1=65\), \(V_2=71\) |
| E.g. \(78 = 1.44a - 0.24b\) and \(71 = 1.44a + 0.04b \Rightarrow 7 = -0.28b \Rightarrow b = ...\) | M1 | Correct method to solve; must show working (matrix method requires inverse) |
| \(a=50, b=-25 \Rightarrow V_n = 50(1.2)^n - 25(-0.2)^n\) | A1* | Correct with no errors seen; fractions instead of decimals acceptable |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(50(1.2)^N > 10^6 \Rightarrow N = ...\) | M1 | Recognises \((-0.2)^n\) negligible; attempts to solve or uses trial and improvement around \(N=55\) |
| \(\Rightarrow N = 55\) i.e. month 55 | A1 | Correct answer implies both marks |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V_1 = 50\times1.2 - 25\times(-0.2) = 60+5 = 65\) and \(V_2 = 50(1.2)^2 - 25(-0.2)^2 = 72-1=71\) | M1 | Substitutes \(n=1\) and \(n=2\) to verify |
| Deduces true for base cases; assumes true for \(n=k\) and \(n=k+1\) | A1 | Must include two successive cases assumed true |
| \(V_{k+2} = V_{k+1} + 0.24V_k = 50(1.2)^{k+1}-25(-0.2)^{k+1}+0.24\big(50(1.2)^k-25(-0.2)^k\big)\) | B1ft | Substitutes formula for \(k\) and \(k+1\) into recurrence |
| \(=\frac{50}{1.2}(1.2)^{k+2}-\frac{25}{-0.2}(-0.2)^{k+2}+\frac{12}{1.2^2}(1.2)^{k+2}-\frac{6}{(-0.2)^2}(-0.2)^{k+2}\) | M1 | Rearranges to form \(a(1.2)^{k+2}+b(-0.2)^{k+2}\) |
| So \(V_{k+2} = 50(1.2)^{k+2}-25(-0.2)^{k+2}\); correct inductive conclusion | A1* | Must include: true for \(n=1,2\); if true for \(n=k\) and \(n=k+1\) then true for \(n=k+2\); hence true for all \(n\in\mathbb{N}\) |
| (5) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V_{n+2} = V_{n+1} + kV_n$ | B1 | Correct expression for model using information given |
| | **(1)** | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda^2 - \lambda - 0.24 = 0 \Rightarrow \lambda = ...(1.2, -0.2)$ | M1 | Forms and solves auxiliary equation with $k=0.24$ |
| $V_n = a(1.2)^n + b(-0.2)^n$ | A1 | Correct closed form; not a follow-through mark |
| $65 = a(1.2)^1 + b(-0.2)^1$ **and** $71 = a(1.2)^2 + b(-0.2)^2$ | B1ft | Applies initial conditions $V_1=65$, $V_2=71$ |
| E.g. $78 = 1.44a - 0.24b$ and $71 = 1.44a + 0.04b \Rightarrow 7 = -0.28b \Rightarrow b = ...$ | M1 | Correct method to solve; must show working (matrix method requires inverse) |
| $a=50, b=-25 \Rightarrow V_n = 50(1.2)^n - 25(-0.2)^n$ | A1* | Correct with no errors seen; fractions instead of decimals acceptable |
| | **(5)** | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50(1.2)^N > 10^6 \Rightarrow N = ...$ | M1 | Recognises $(-0.2)^n$ negligible; attempts to solve or uses trial and improvement around $N=55$ |
| $\Rightarrow N = 55$ i.e. month 55 | A1 | Correct answer implies both marks |
| | **(2)** | |
### Alternative (Induction) for Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V_1 = 50\times1.2 - 25\times(-0.2) = 60+5 = 65$ and $V_2 = 50(1.2)^2 - 25(-0.2)^2 = 72-1=71$ | M1 | Substitutes $n=1$ and $n=2$ to verify |
| Deduces true for base cases; assumes true for $n=k$ and $n=k+1$ | A1 | Must include two successive cases assumed true |
| $V_{k+2} = V_{k+1} + 0.24V_k = 50(1.2)^{k+1}-25(-0.2)^{k+1}+0.24\big(50(1.2)^k-25(-0.2)^k\big)$ | B1ft | Substitutes formula for $k$ and $k+1$ into recurrence |
| $=\frac{50}{1.2}(1.2)^{k+2}-\frac{25}{-0.2}(-0.2)^{k+2}+\frac{12}{1.2^2}(1.2)^{k+2}-\frac{6}{(-0.2)^2}(-0.2)^{k+2}$ | M1 | Rearranges to form $a(1.2)^{k+2}+b(-0.2)^{k+2}$ |
| So $V_{k+2} = 50(1.2)^{k+2}-25(-0.2)^{k+2}$; correct inductive conclusion | A1* | Must include: true for $n=1,2$; if true for $n=k$ and $n=k+1$ then true for $n=k+2$; hence true for all $n\in\mathbb{N}$ |
| | **(5)** | |
---\begin{enumerate}
\item The number of visits to a website, in any particular month, is modelled as the number of visits received in the previous month plus $k$ times the number of visits received in the month before that, where $k$ is a positive constant.
\end{enumerate}
Given that $V _ { n }$ is the number of visits to the website in month $n$,\\
(a) write down a general recurrence relation for $V _ { n + 2 }$ in terms of $V _ { n + 1 } , V _ { n }$ and $k$.
For a particular website you are given that
\begin{itemize}
\item $k = 0.24$
\item In month 1 , there were 65 visits to the website.
\item In month 2 , there were 71 visits to the website.\\
(b) Show that
\end{itemize}
$$V _ { n } = 50 ( 1.2 ) ^ { n } - 25 ( - 0.2 ) ^ { n }$$
This model predicts that the number of visits to this website will exceed one million for the first time in month $N$.\\
(c) Find the value of $N$.
\hfill \mbox{\textit{Edexcel FP2 2019 Q3 [8]}}