OCR Further Pure Core 2 2020 November — Question 3 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions for summation
DifficultyStandard +0.8 This is a Further Maths question requiring partial fractions decomposition, telescoping series summation, and careful algebraic manipulation to reach a specific closed form. While the techniques are standard for Further Pure, the multi-step process (factorising, decomposing, recognising the telescoping pattern, and simplifying to the exact form given) requires more sophistication than typical A-level questions. Part (b) is straightforward given part (a).
Spec4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series

3 In this question you must show detailed reasoning.
  1. Use partial fractions to show that \(\sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }\).
  2. Write down the value of \(\lim _ { n \rightarrow \infty } \left( \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } \right)\).

3 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Use partial fractions to show that $\sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }$.
\item Write down the value of $\lim _ { n \rightarrow \infty } \left( \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } \right)$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q3 [6]}}