Question 3 - A-Level Maths

CAIE FP1 2018 November — Question 3 8 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2018
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and series, recurrence and convergence
Type	Non-linear or complex iterative formula convergence
Difficulty	Challenging +1.2 This is a standard Further Maths recurrence relation question requiring induction proof and monotonicity analysis. Part (i) follows a heavily guided approach ('by considering 3 - u_{n+1}') making the algebraic manipulation straightforward. Part (ii) is routine comparison of consecutive terms. While it requires careful algebra and understanding of sequences, the techniques are well-practiced in FP1 with no novel insight needed—moderately above average difficulty due to the Further Maths context and multi-step reasoning.
Spec	4.01a Mathematical induction: construct proofs

The sequence of positive numbers $u_1$, $u_2$, $u_3$, $\ldots$ is such that $u_1 < 3$ and, for $n \geqslant 1$, $$u_{n+1} = \frac{4u_n + 9}{u_n + 4}.$$

By considering $3 - u_{n+1}$, or otherwise, prove by mathematical induction that $u_n < 3$ for all positive integers $n$. [5]
Show that $u_{n+1} > u_n$ for $n \geqslant 1$. [3]

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
3(i)	u <3 ( given )
1	B1	States base case.

Assume that u <3

Answer	Marks	Guidance
k	B1	States inductive hypothesis.

Then

4u +9 −u +3

3−u =3− k = k >0 ⇒ u <3

k+1 u +4 u +4 k+1

Answer	Marks
k k	M1 A1

Hence, by induction, u <3 for all n(cid:46)1.

Answer	Marks	Guidance
n	B1	States conclusion.

5

Answer	Marks
3(ii)	4u +9 −u2 +9

u −u = n −u = n

n+1 n u +4 n u +4

Answer	Marks	Guidance
n n	M1 A1	Considers u −u .

n+1 n

So u <3⇒u −u >0.

Answer	Marks	Guidance
n n+1 n	B1	Uses u <3

n

3

Answer	Marks	Guidance
Question	Answer	Marks

Question 3:
--- 3(i) ---
3(i) | u <3 ( given )
1 | B1 | States base case.
Assume that u <3
k | B1 | States inductive hypothesis.
Then
4u +9 −u +3
3−u =3− k = k >0 ⇒ u <3
k+1 u +4 u +4 k+1
k k | M1 A1
Hence, by induction, u <3 for all n(cid:46)1.
n | B1 | States conclusion.
5
--- 3(ii) ---
3(ii) | 4u +9 −u2 +9
u −u = n −u = n
n+1 n u +4 n u +4
n n | M1 A1 | Considers u −u .
n+1 n
So u <3⇒u −u >0.
n n+1 n | B1 | Uses u <3
n
3
Question | Answer | Marks | Guidance

Show LaTeX source

The sequence of positive numbers $u_1$, $u_2$, $u_3$, $\ldots$ is such that $u_1 < 3$ and, for $n \geqslant 1$,
$$u_{n+1} = \frac{4u_n + 9}{u_n + 4}.$$

\begin{enumerate}[label=(\roman*)]
\item By considering $3 - u_{n+1}$, or otherwise, prove by mathematical induction that $u_n < 3$ for all positive integers $n$. [5]

\item Show that $u_{n+1} > u_n$ for $n \geqslant 1$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q3 [8]}}

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