| Exam Board | CAIE |
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| Module | FP1 (Further Pure Mathematics 1) |
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| Year | 2006 |
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| Session | November |
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| Topic | Sequences and series, recurrence and convergence |
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3 Verify that if
$$v _ { n } = n ( n + 1 ) ( n + 2 ) \ldots ( n + m )$$
then
$$v _ { n + 1 } - v _ { n } = ( m + 1 ) ( n + 1 ) ( n + 2 ) \ldots ( n + m ) .$$
Given now that
$$u _ { n } = ( n + 1 ) ( n + 2 ) \ldots ( n + m ) ,$$
find \(\sum _ { n = 1 } ^ { N } u _ { n }\) in terms of \(m\) and \(N\).