2006 TYS

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]