AQA FP2 2011 January — Question 7 7 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJanuary
Marks7
PaperDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.8 This is a structured two-part induction proof requiring algebraic manipulation to establish a recurrence relation, then applying it in the inductive step. Part (a) guides students through finding the key relationship (a=7), making the induction more accessible than if unscaffolded. The divisibility proof itself is standard Further Maths fare, but the preliminary algebraic work and the need to recognize how to use the recurrence relation elevates it slightly above a routine induction question.
Spec4.01a Mathematical induction: construct proofs

7
  1. Given that $$\mathrm { f } ( k ) = 12 ^ { k } + 2 \times 5 ^ { k - 1 }$$ show that $$\mathrm { f } ( k + 1 ) - 5 \mathrm { f } ( k ) = a \times 12 ^ { k }$$ where \(a\) is an integer.
  2. Prove by induction that \(12 ^ { n } + 2 \times 5 ^ { n - 1 }\) is divisible by 7 for all integers \(n \geqslant 1\).

7
\begin{enumerate}[label=(\alph*)]
\item Given that

$$\mathrm { f } ( k ) = 12 ^ { k } + 2 \times 5 ^ { k - 1 }$$

show that

$$\mathrm { f } ( k + 1 ) - 5 \mathrm { f } ( k ) = a \times 12 ^ { k }$$

where $a$ is an integer.
\item Prove by induction that $12 ^ { n } + 2 \times 5 ^ { n - 1 }$ is divisible by 7 for all integers $n \geqslant 1$.
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2011 Q7 [7]}}