| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Standard +0.8 This FP2 question requires understanding recurrence relations in context, computing terms, and deriving the closed-form Binet formula for Fibonacci-type sequences. Part (c) demands solving the characteristic equation and applying initial conditions—standard FP2 technique but requiring careful algebraic manipulation across multiple steps. More challenging than typical C1/C2 but routine for Further Maths students who've studied recurrence relations. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method8.01a Recurrence relations: general sequences, closed form and recurrence8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_1 = 1\) as there is only one way to go up one step | B1 | Need to see explanation for \(u_1 = 1\) |
| \(u_2 = 2\) as there are two ways: one step then one step, or two steps | B1 | Need to see explanation for \(u_2 = 2\) with the two ways spelled out |
| If first move is one step: climb other \((n-1)\) steps in \(u_{n-1}\) ways; if first move is two steps: climb other \((n-2)\) steps in \(u_{n-2}\) ways, so \(u_n = u_{n-1} + u_{n-2}\) | B1 | First move can be one or two steps with clear explanation of iterative expression |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Sequence \(1, 2, 3, 5, 8, 13, 21, 34, \ldots\) so \(34\) ways of climbing \(8\) steps | B1 | The answer is enough for this mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_n = u_{n-1} + u_{n-2}\) gives \(\lambda^2 = \lambda + 1\) | M1 | Obtains this characteristic equation |
| Roots \(\dfrac{1 \pm \sqrt{5}}{2}\) | A1 | Solves quadratic giving exact answers |
| General form \(A\left(\dfrac{1+\sqrt{5}}{2}\right)^n + B\left(\dfrac{1-\sqrt{5}}{2}\right)^n\) | M1 | Obtains a general form |
| Uses initial conditions to find \(A\) and \(B\), reaching two equations in \(A\) and \(B\); conditions are \(A(1+\sqrt{5}) + B(1-\sqrt{5}) = 2\) and \(A(3+\sqrt{5}) + B(3-\sqrt{5}) = 4\) | M1 | Use initial conditions to obtain two equations (slips allowed) |
| \(A = \dfrac{1+\sqrt{5}}{2\sqrt{5}}\), \(B = -\dfrac{1-\sqrt{5}}{2\sqrt{5}}\); for \(n=400\): \(\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^{401} - \left(\dfrac{1-\sqrt{5}}{2}\right)^{401}\right]\) | A1* | Must see exact correct values for \(A\) and \(B\) and conclusion given for \(n = 400\) |
# Question 8:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_1 = 1$ as there is only one way to go up one step | B1 | Need to see explanation for $u_1 = 1$ |
| $u_2 = 2$ as there are two ways: one step then one step, or two steps | B1 | Need to see explanation for $u_2 = 2$ with the two ways spelled out |
| If first move is one step: climb other $(n-1)$ steps in $u_{n-1}$ ways; if first move is two steps: climb other $(n-2)$ steps in $u_{n-2}$ ways, so $u_n = u_{n-1} + u_{n-2}$ | B1 | First move can be one or two steps with clear explanation of iterative expression |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sequence $1, 2, 3, 5, 8, 13, 21, 34, \ldots$ so $34$ ways of climbing $8$ steps | B1 | The answer is enough for this mark |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_n = u_{n-1} + u_{n-2}$ gives $\lambda^2 = \lambda + 1$ | M1 | Obtains this characteristic equation |
| Roots $\dfrac{1 \pm \sqrt{5}}{2}$ | A1 | Solves quadratic giving exact answers |
| General form $A\left(\dfrac{1+\sqrt{5}}{2}\right)^n + B\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ | M1 | Obtains a general form |
| Uses initial conditions to find $A$ and $B$, reaching two equations in $A$ and $B$; conditions are $A(1+\sqrt{5}) + B(1-\sqrt{5}) = 2$ and $A(3+\sqrt{5}) + B(3-\sqrt{5}) = 4$ | M1 | Use initial conditions to obtain two equations (slips allowed) |
| $A = \dfrac{1+\sqrt{5}}{2\sqrt{5}}$, $B = -\dfrac{1-\sqrt{5}}{2\sqrt{5}}$; for $n=400$: $\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^{401} - \left(\dfrac{1-\sqrt{5}}{2}\right)^{401}\right]$ | A1* | Must see exact correct values for $A$ and $B$ and conclusion given for $n = 400$ |\begin{enumerate}
\item A staircase has $n$ steps. A tourist moves from the bottom (step zero) to the top (step $n$ ). At each move up the staircase she can go up either one step or two steps, and her overall climb up the staircase is a combination of such moves.
\end{enumerate}
If $u _ { n }$ is the number of ways that the tourist can climb up a staircase with $n$ steps,\\
(a) explain why $u _ { n }$ satisfies the recurrence relation
$$u _ { n } = u _ { n - 1 } + u _ { n - 2 } , \text { with } u _ { 1 } = 1 \text { and } u _ { 2 } = 2$$
(b) Find the number of ways in which she can climb up a staircase when there are eight steps.
A staircase at a certain tourist attraction has 400 steps.\\
(c) Show that the number of ways in which she could climb up to the top of this staircase is given by
$$\frac { 1 } { \sqrt { 5 } } \left[ \left( \frac { 1 + \sqrt { 5 } } { 2 } \right) ^ { 401 } - \left( \frac { 1 - \sqrt { 5 } } { 2 } \right) ^ { 401 } \right]$$
\hfill \mbox{\textit{Edexcel FP2 Q8 [9]}}