OCR MEI FP1 2005 January — Question 2 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyModerate -0.5 This is a straightforward application of the method of differences with the identity explicitly provided. Part (i) is trivial algebra (common denominator), and part (ii) requires only mechanical telescoping of the given form. While method of differences is a Further Maths topic, the question involves no problem-solving or insight since the decomposition is given, making it easier than average overall.
Spec4.06b Method of differences: telescoping series

2
  1. Show that \(\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { 1 } { ( r + 1 ) ( r + 2 ) }\).
  2. Hence use the method of differences to find the sum of the series $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }$$

2 (i) Show that $\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { 1 } { ( r + 1 ) ( r + 2 ) }$.\\
(ii) Hence use the method of differences to find the sum of the series

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }$$

\hfill \mbox{\textit{OCR MEI FP1 2005 Q2 [6]}}