AQA FP2 2013 June — Question 4 7 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyChallenging +1.2 This is a standard method of differences question with the identity provided. Part (a) is routine algebraic verification (expand and simplify), and part (b) applies the telescoping sum technique directly. While it requires careful algebraic manipulation and understanding of the method, it's a textbook application of a Further Maths technique with no novel insight needed.
Spec4.06b Method of differences: telescoping series

4
  1. Given that \(\mathrm { f } ( r ) = r ^ { 2 } \left( 2 r ^ { 2 } - 1 \right)\), show that $$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = ( 2 r - 1 ) ^ { 3 }$$
  2. Use the method of differences to show that $$\sum _ { r = n + 1 } ^ { 2 n } ( 2 r - 1 ) ^ { 3 } = 3 n ^ { 2 } \left( 10 n ^ { 2 } - 1 \right)$$ (4 marks)

4
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ( r ) = r ^ { 2 } \left( 2 r ^ { 2 } - 1 \right)$, show that

$$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = ( 2 r - 1 ) ^ { 3 }$$
\item Use the method of differences to show that

$$\sum _ { r = n + 1 } ^ { 2 n } ( 2 r - 1 ) ^ { 3 } = 3 n ^ { 2 } \left( 10 n ^ { 2 } - 1 \right)$$

(4 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2013 Q4 [7]}}