AQA FP2 2011 June — Question 6 8 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.3 Part (a) is straightforward algebraic expansion requiring no insight. Part (b) is a standard textbook induction proof with a given formula—students follow the mechanical steps of base case, assumption, and inductive step using the algebraic identity from (a). While induction is an FP2 topic, this is a routine application with no novel problem-solving required, making it slightly easier than average overall.
Spec4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3

  1. Show that $$(k + 1)(4(k + 1)^2 - 1) = 4k^3 + 12k^2 + 11k + 3$$ [2 marks]
  2. Prove by induction that, for all integers \(n \geqslant 1\), $$1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2 = \frac{1}{3}n(4n^2 - 1)$$ [6 marks]

\begin{enumerate}[label=(\alph*)]
\item Show that
$$(k + 1)(4(k + 1)^2 - 1) = 4k^3 + 12k^2 + 11k + 3$$ [2 marks]

\item Prove by induction that, for all integers $n \geqslant 1$,
$$1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2 = \frac{1}{3}n(4n^2 - 1)$$ [6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2011 Q6 [8]}}