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

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

$\sum_{n=1}^{N}u_n=\frac{(N+1)(N+2)\cdots(N+1+m)-1\cdot2\cdot3\cdots(m+1)}{m+1}$

N2006 P1 Q3 (FM)

N2006 P1 Q3 (FM)

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