| Exam Board | OCR MEI |
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| Module | FP1 (Further Pure Mathematics 1) |
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| Year | 2014 |
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| Session | June |
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| Topic | Sequences and series, recurrence and convergence |
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4 Use the identity \(\frac { 1 } { 2 r + 3 } - \frac { 1 } { 2 r + 5 } \equiv \frac { 2 } { ( 2 r + 3 ) ( 2 r + 5 ) }\) and the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 3 ) ( 2 r + 5 ) }\), expressing your answer as a single fraction.