Partial Fractions and Telescoping Series

A question is this type if and only if it involves expressing a term in partial fractions and using telescoping to find a sum or sum to infinity.

4 questions · Challenging +1.1

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CAIE Further Paper 1 2020 June Q3
9 marks Challenging +1.2
3 Let \(S _ { n } = 2 ^ { 2 } + 6 ^ { 2 } + 10 ^ { 2 } + \ldots + ( 4 n - 2 ) ^ { 2 }\).
  1. Use standard results from the List of Formulae (MF19) to show that \(S _ { n } = \frac { 4 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)\).
  2. Express \(\frac { \mathrm { n } } { \mathrm { S } _ { \mathrm { n } } }\) in partial fractions and find \(\sum _ { \mathrm { n } = 1 } ^ { \mathrm { N } } \frac { \mathrm { n } } { \mathrm { S } _ { \mathrm { n } } }\) in terms of \(N\).
  3. Deduce the value of \(\sum _ { n = 1 } ^ { \infty } \frac { n } { S _ { n } }\).
CAIE Further Paper 1 2022 November Q1
7 marks Standard +0.3
1
  1. Use the list of formulae (MF19) to find \(\sum _ { r = 1 } ^ { n } r ( r + 2 )\) in terms of \(n\), simplifying your answer.
  2. Express \(\frac { 1 } { r ( r + 2 ) }\) in partial fractions and hence find \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { \mathrm { r } ( \mathrm { r } + 2 ) }\) in terms of \(n\).
  3. Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 2 ) }\).
CAIE Further Paper 1 2024 November Q5
9 marks Challenging +1.2
5 It is given that \(S _ { n } = \sum _ { r = 1 } ^ { n } u _ { r }\), where \(u _ { r } = x ^ { \mathrm { f } ( r ) } - x ^ { \mathrm { f } ( r + 1 ) }\) and \(x > 0\).
  1. Find \(S _ { n }\) in terms of \(n , x\) and the function f .
  2. Given that \(\mathrm { f } ( r ) = \ln r\), find the set of values of \(x\) for which the infinite series $$u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$$ is convergent and give the sum to infinity when this exists.
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  3. Given instead that \(\mathrm { f } ( r ) = 2 \log _ { x } r\) where \(x \neq 1\), use standard results from the List of formulae (MF19) to find \(\sum _ { n = 1 } ^ { N } S _ { n }\) in terms of \(N\). Fully factorise your answer.
CAIE FP1 2010 June Q2
5 marks Challenging +1.8
2 By considering the identity $$\cos [ ( 2 n - 1 ) \alpha ] - \cos [ ( 2 n + 1 ) \alpha ] \equiv 2 \sin \alpha \sin 2 n \alpha$$ show that if \(\alpha\) is not an integer multiple of \(\pi\) then $$\sum _ { n = 1 } ^ { N } \sin ( 2 n \alpha ) = \frac { 1 } { 2 } \cot \alpha - \frac { 1 } { 2 } \operatorname { cosec } \alpha \cos [ ( 2 N + 1 ) \alpha ]$$ Deduce that the infinite series $$\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 2 } { 3 } n \pi \right)$$ does not converge.