| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.2 This is a standard proof by induction for a recurrence relation with a given closed form. While it requires careful algebraic manipulation with factorials and powers, the structure is routine: verify base cases, assume for n and n+1, then prove for n+2 by substituting into the recurrence relation. The factorial and power manipulation is straightforward for Further Maths students, making this a moderately above-average difficulty question but not requiring novel insight. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(n=1: u_1=(-3)^1\times1!=-3\); \(n=2: u_2=(-3)^2\times2!=9\times2=18\); hence true for \(n=1\) and \(n=2\) | B1 | Checks closed form works for \(n=1\) and \(n=2\) |
| Assume true for \(n=k\) and \(n=k+1\), so \(u_k=(-3)^k k!\) and \(u_{k+1}=(-3)^{k+1}(k+1)!\) | M1 | Makes inductive assumption. Must be clear it is the closed forms being assumed |
| Then \(u_{k+2}=9(k+1)^2((-3)^k k!)-3((-3)^{k+1}(k+1)!)\) | M1 | Substitutes expressions for \(n=k\) and \(n=k+1\) into recurrence formula |
| \(=(-3)^k k!\left[9(k+1)^2-3(-3)(k+1)\right]\) | M1 | Takes out common factor of at least \((-3)^k k!\); treatment of \((-3)\) must be correct |
| \(=(-3)^k k!\left[9(k+1)(k+1+1)\right]=(-3)^k\times(-3)^2\times(k+1)(k+2)k!\) \(=(-3)^{k+2}(k+2)!\) | A1 | Simplifies correctly to required form for assumed true values |
| Hence if true for \(n=k\) and \(n=k+1\) then true for \(n=k+2\). As also true for \(n=1\) and \(n=2\), then true for all \(n\in\mathbb{N}\) by mathematical induction | A1 | Correct conclusion: conveys 1) true for \(n=1,2\); 2) true for two successive cases implies true for next; 3) true for all positive \(n\) |
| (6) |
# Question 6:
| Working | Mark | Guidance |
|---------|------|----------|
| $n=1: u_1=(-3)^1\times1!=-3$; $n=2: u_2=(-3)^2\times2!=9\times2=18$; hence true for $n=1$ and $n=2$ | **B1** | Checks closed form works for $n=1$ and $n=2$ |
| Assume true for $n=k$ and $n=k+1$, so $u_k=(-3)^k k!$ and $u_{k+1}=(-3)^{k+1}(k+1)!$ | **M1** | Makes inductive assumption. Must be clear it is the closed forms being assumed |
| Then $u_{k+2}=9(k+1)^2((-3)^k k!)-3((-3)^{k+1}(k+1)!)$ | **M1** | Substitutes expressions for $n=k$ and $n=k+1$ into recurrence formula |
| $=(-3)^k k!\left[9(k+1)^2-3(-3)(k+1)\right]$ | **M1** | Takes out common factor of at least $(-3)^k k!$; treatment of $(-3)$ must be correct |
| $=(-3)^k k!\left[9(k+1)(k+1+1)\right]=(-3)^k\times(-3)^2\times(k+1)(k+2)k!$ $=(-3)^{k+2}(k+2)!$ | **A1** | Simplifies correctly to required form for assumed true values |
| Hence if true for $n=k$ and $n=k+1$ then true for $n=k+2$. As also true for $n=1$ and $n=2$, then true for all $n\in\mathbb{N}$ by mathematical induction | **A1** | Correct conclusion: conveys 1) true for $n=1,2$; 2) true for two successive cases implies true for next; 3) true for all positive $n$ |
| | **(6)** | |\begin{enumerate}
\item A recurrence system is defined by
\end{enumerate}
$$\begin{aligned}
u _ { n + 2 } & = 9 ( n + 1 ) ^ { 2 } u _ { n } - 3 u _ { n + 1 } \quad n \geqslant 1 \\
u _ { 1 } & = - 3 , u _ { 2 } = 18
\end{aligned}$$
Prove by induction that, for $n \in \mathbb { N }$,
$$u _ { n } = ( - 3 ) ^ { n } n !$$
\hfill \mbox{\textit{Edexcel FP2 2021 Q6 [6]}}