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

Question

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]

Accepted Answer

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

N2006 P1 Q3 (FM)

N2006 P1 Q3 (FM)

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