| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Challenging +1.2 This is a standard FP2 recurrence relation question with three routine parts: (a) explaining a given recurrence relation from a word problem (straightforward interpretation), (b) proof by induction following a provided formula (mechanical application of induction steps), and (c) model evaluation (simple substitution and comparison). While it involves Further Maths content, each part follows predictable patterns with no novel problem-solving required, making it slightly above average difficulty due to the multi-step nature and induction component. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Two of: \(U_1=25\); 20% leave so \(0.8U_n\) remain; \(20(n+1)\) new subscribers added at end of week \(n\) | M1 | At least two aspects explained |
| \(U_{n+1}=0.8U_n+20(n+1),\quad U_1=25\) | A1 | All three aspects combined correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=1\Rightarrow U_1=325\times1+100\times1-400=25\) ✓ | B1 | Base case verified |
| \(U_{k+1}=0.8U_k+20k+20=0.8\!\left(325\!\left(\tfrac{4}{5}\right)^{k-1}+100k-400\right)+20k+20\) | M1 | Inductive assumption made and closed form substituted |
| \(=325\times\tfrac{4}{5}\times\left(\tfrac{4}{5}\right)^{k-1}+80k-320+20k+20 = 325\times\left(\tfrac{4}{5}\right)^k+100k-300\) | M1 | Simplifies combining powers of \(\tfrac{4}{5}\) into one term |
| \(=325\times\left(\tfrac{4}{5}\right)^k+100(k+1)-400\) | A1 | Correct form with \(k+1\) in linear term |
| Conclusion: true for \(n=k\Rightarrow\) true for \(n=k+1\); true for \(n=1\); therefore true for all \(n\geqslant1\) | A1 | All bold conclusion statements required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt at \(U_{24}=325\times0.8^{23}+2400-400=2001.9\ldots\), or \(U_{25}=2101.5\ldots\), or \(U_{26}=2201.2\ldots\), or \(U_{27}=2300.98\ldots\) | M1 | Attempt at relevant week value |
| Correct value for weeks 24–27; compares with 1800 after 6 months; e.g. model overestimates by 400 people, therefore not a very good model | A1 | Valid comparison and conclusion |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Two of: $U_1=25$; 20% leave so $0.8U_n$ remain; $20(n+1)$ new subscribers added at end of week $n$ | M1 | At least two aspects explained |
| $U_{n+1}=0.8U_n+20(n+1),\quad U_1=25$ | A1 | All three aspects combined correctly |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1\Rightarrow U_1=325\times1+100\times1-400=25$ ✓ | B1 | Base case verified |
| $U_{k+1}=0.8U_k+20k+20=0.8\!\left(325\!\left(\tfrac{4}{5}\right)^{k-1}+100k-400\right)+20k+20$ | M1 | Inductive assumption made and closed form substituted |
| $=325\times\tfrac{4}{5}\times\left(\tfrac{4}{5}\right)^{k-1}+80k-320+20k+20 = 325\times\left(\tfrac{4}{5}\right)^k+100k-300$ | M1 | Simplifies combining powers of $\tfrac{4}{5}$ into one term |
| $=325\times\left(\tfrac{4}{5}\right)^k+100(k+1)-400$ | A1 | Correct form with $k+1$ in linear term |
| Conclusion: true for $n=k\Rightarrow$ true for $n=k+1$; true for $n=1$; therefore true for all $n\geqslant1$ | A1 | All bold conclusion statements required |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt at $U_{24}=325\times0.8^{23}+2400-400=2001.9\ldots$, or $U_{25}=2101.5\ldots$, or $U_{26}=2201.2\ldots$, or $U_{27}=2300.98\ldots$ | M1 | Attempt at relevant week value |
| Correct value for weeks 24–27; compares with 1800 after 6 months; e.g. model overestimates by 400 people, therefore not a very good model | A1 | Valid comparison and conclusion |\begin{enumerate}
\item In a model for the number of subscribers to a new social media channel it is assumed that
\end{enumerate}
\begin{itemize}
\item each week $20 \%$ of the subscribers at the start of the week cancel their subscriptions
\item between the start and end of week $n$ the channel gains $20 n$ new subscribers
\end{itemize}
Given that at the end of week 1 there were 25 subscribers,\\
(a) explain why the number of subscribers at the end of week $n , U _ { n }$, is modelled by the recurrence relation
$$U _ { 1 } = 25 \quad U _ { n + 1 } = 0.8 U _ { n } + 20 ( n + 1 ) \quad n = 1,2,3 , \ldots$$
(b) Prove by induction that for $n \geqslant 1$
$$U _ { n } = 325 \left( \frac { 4 } { 5 } \right) ^ { n - 1 } + 100 n - 400$$
Given that 6 months after starting the channel there were approximately 1800 subscribers,\\
(c) evaluate the model in the light of this information.
\hfill \mbox{\textit{Edexcel FP2 2023 Q3 [9]}}