AQA FP2 2012 June — Question 7 9 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
TopicProof by induction
TypeProve summation with fractions
DifficultyStandard +0.8 This is a Further Maths induction proof with algebraic fractions requiring careful manipulation to show the inductive step works, plus a follow-up inequality question. The algebraic complexity (dealing with n²(n+1)² denominators) and the need to correctly handle the k+1 case makes this moderately challenging, though it follows standard induction structure.
Spec4.01a Mathematical induction: construct proofs

7
  1. Prove by induction that, for all integers \(n \geqslant 1\), $$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots + \frac { 2 n + 1 } { n ^ { 2 } ( n + 1 ) ^ { 2 } } = 1 - \frac { 1 } { ( n + 1 ) ^ { 2 } }$$
  2. Find the smallest integer \(n\) for which the sum of the series differs from 1 by less than \(10 ^ { - 5 }\).

7
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, for all integers $n \geqslant 1$,

$$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots + \frac { 2 n + 1 } { n ^ { 2 } ( n + 1 ) ^ { 2 } } = 1 - \frac { 1 } { ( n + 1 ) ^ { 2 } }$$
\item Find the smallest integer $n$ for which the sum of the series differs from 1 by less than $10 ^ { - 5 }$.
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2012 Q7 [9]}}