| Exam Board | OCR |
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| Module | FP1 (Further Pure Mathematics 1) |
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| Year | 2005 |
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| Session | June |
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| Topic | Proof by induction |
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1 Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 2 r + 1 \right) = n \left( 2 n ^ { 2 } + 4 n + 3 \right)$$