CAIE FP1 2006 November — Question 3 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionNovember
Marks5
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeFactorial or product method of differences
DifficultyChallenging +1.2 This is a structured Further Maths question on telescoping series with factorial-type products. The first part requires straightforward algebraic verification by factoring, while the second part applies the method of differences—a standard FP1 technique. Though it involves products and requires recognizing the telescoping pattern, the question provides explicit guidance through the verification step, making it more accessible than unscaffolded proof problems. It's harder than routine A-level due to the Further Maths content and multi-step reasoning, but remains a textbook application of a known method.
Spec4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series

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 \(N\).

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 $N$.

\hfill \mbox{\textit{CAIE FP1 2006 Q3 [5]}}