OCR FP1 2008 June — Question 3 6 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFactorial or product method of differences
DifficultyStandard +0.3 This is a straightforward method of differences question requiring algebraic manipulation to verify the given identity, then telescoping sum application. The factorial manipulation in part (i) is routine, and part (ii) follows directly by recognizing the telescoping pattern. Slightly easier than average as it's a standard FP1 technique with clear signposting.
Spec4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series

3
  1. Show that \(\frac { 1 } { r ! } - \frac { 1 } { ( r + 1 ) ! } = \frac { r } { ( r + 1 ) ! }\).
  2. Hence find an expression, in terms of \(n\), for $$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n } { ( n + 1 ) ! }$$

3 (i) Show that $\frac { 1 } { r ! } - \frac { 1 } { ( r + 1 ) ! } = \frac { r } { ( r + 1 ) ! }$.\\
(ii) Hence find an expression, in terms of $n$, for

$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n } { ( n + 1 ) ! }$$

\hfill \mbox{\textit{OCR FP1 2008 Q3 [6]}}