Edexcel FP2 2016 June — Question 2 7 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2016
SessionJune
Marks7
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeMethod of differences with given identity
DifficultyStandard +0.8 This is a Further Maths FP2 question requiring algebraic manipulation to verify an identity, then applying method of differences to sum a series. While the method of differences is a standard FP2 technique, the algebraic complexity of tracking telescoping terms and simplifying to the given form with unknown constants requires careful bookwork and is more demanding than typical A-level questions. The multi-step nature and need to find constants a and b elevates this above average difficulty.
Spec4.06b Method of differences: telescoping series

2. (a) Show that, for \(r > 0\) $$r - 3 + \frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) }$$ (b) Hence prove, using the method of differences, that $$\sum _ { r = 1 } ^ { n } \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) } = \frac { n \left( n ^ { 2 } + a n + b \right) } { 2 ( n + 2 ) }$$ where \(a\) and \(b\) are constants to be found.

2. (a) Show that, for $r > 0$

$$r - 3 + \frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) }$$

(b) Hence prove, using the method of differences, that

$$\sum _ { r = 1 } ^ { n } \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) } = \frac { n \left( n ^ { 2 } + a n + b \right) } { 2 ( n + 2 ) }$$

where $a$ and $b$ are constants to be found.

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