2006 A-Level H2 Mathematics

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Question 3Sequence and Series5 marks

Verify that if

${{v}_{n}}=n\left( n+1 \right)\left( n+2 \right)...\left( n+m \right)$.

then

${{v}_{n+1}}-{{v}_{n}}=\left( m+1 \right)\left( n+1 \right)\left( n+2 \right)...\left( n+m \right)$.[2]

Given now that

${{u}_{n}}=\left( n+1 \right)\left( n+2 \right)...\left( n+m \right)$,

find $\sum\limits_{n=1}^{N}{{{u}_{n}}}$ in terms of $m$ and $N$.[3]

Final Answer:

${\frac{1}{m+1}[(N+1)(N+2)(N+3)...(N+1+m)-(1)(2)(3)...(m+1)]}$