Edexcel F1 2021 January — Question 9 12 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2021
SessionJanuary
Marks12
PaperDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyStandard +0.3 This is a standard two-part induction question from Further Pure 1. Part (i) involves proving a closed form for a recurrence relation using straightforward algebraic manipulation, while part (ii) is a routine divisibility proof. Both follow textbook templates with no novel insights required, making this slightly easier than average even for Further Maths students.
Spec4.01a Mathematical induction: construct proofs

9. (i) A sequence of numbers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by $$u _ { n + 1 } = \frac { 1 } { 3 } \left( 2 u _ { n } - 1 \right) \quad u _ { 1 } = 1$$ Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\) $$u _ { n } = 3 \left( \frac { 2 } { 3 } \right) ^ { n } - 1$$ (ii) \(\mathrm { f } ( n ) = 2 ^ { n + 2 } + 3 ^ { 2 n + 1 }\) Prove by induction that, for \(n \in \mathbb { Z } ^ { + } , \mathrm { f } ( n )\) is a multiple of 7
VIXV SIHIANI III IM IONOOVIAV SIHI NI JYHAM ION OOVI4V SIHI NI JLIYM ION OO

9. (i) A sequence of numbers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by

$$u _ { n + 1 } = \frac { 1 } { 3 } \left( 2 u _ { n } - 1 \right) \quad u _ { 1 } = 1$$

Prove by induction that, for $n \in \mathbb { Z } ^ { + }$

$$u _ { n } = 3 \left( \frac { 2 } { 3 } \right) ^ { n } - 1$$

(ii) $\mathrm { f } ( n ) = 2 ^ { n + 2 } + 3 ^ { 2 n + 1 }$

Prove by induction that, for $n \in \mathbb { Z } ^ { + } , \mathrm { f } ( n )$ is a multiple of 7\\

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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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