Question 3 - A-Level Maths

AQA FP2 2013 June — Question 3 6 marks

Exam Board	AQA
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Marks	6
Paper	Download PDF ↗
Topic	Proof by induction
Type	Prove recurrence relation formula
Difficulty	Standard +0.8 This is a standard FP2 induction proof with a recurrence relation, requiring algebraic manipulation to show P(k)⇒P(k+1). The algebra involves substituting a fraction into another fraction and simplifying, which is moderately demanding but follows a well-practiced template. It's harder than basic C1/C2 induction (sums of series) but routine for Further Maths students who have practiced this question type.
Spec	4.01a Mathematical induction: construct proofs

3 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by $$u _ { 1 } = 2 , \quad u _ { n + 1 } = \frac { 5 u _ { n } - 3 } { 3 u _ { n } - 1 }$$ Prove by induction that, for all integers $n \geqslant 1$, $$u _ { n } = \frac { 3 n + 1 } { 3 n - 1 }$$ (6 marks)

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3 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by

$$u _ { 1 } = 2 , \quad u _ { n + 1 } = \frac { 5 u _ { n } - 3 } { 3 u _ { n } - 1 }$$

Prove by induction that, for all integers $n \geqslant 1$,

$$u _ { n } = \frac { 3 n + 1 } { 3 n - 1 }$$

(6 marks)

\hfill \mbox{\textit{AQA FP2 2013 Q3 [6]}}

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Q1 7 Q2 9 Q3 6 Q4 7 Q5 9 Q6 8 Q7 12 Q8 17