TITLE: Find the GCD of $10$ integers
QUESTION [5 upvotes]: Find the greatest common divisor of the following ten integers: 
  \begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\  2001^3+3\cdot2001^2+2\cdot2001,\  \dots, \\[5pt]
& \dots, \  2008^3+3\cdot2008^2+2\cdot2008,\  2009^3+3\cdot2009^2+2\cdot2009.
\end{align}

How are these numbers related to each other? Is there a formula for this?

REPLY [2 votes]: The sequence  $\,f_n = n(n\!+\!1)(n\!+\!2)\,$ satisfies $\,\nabla^3 f_n = 6,\,$ where $\,\nabla f_n := f_{n+1}-f_n,\, $ therefore $\,d \mid f_n,f_{n+1},f_{n+2},f_{n+3}\,\Rightarrow\,d\mid \nabla^3 f_n = 6.\,$ Conversely $\ 6\mid f_n = 3!{n+2\choose 3}\ $ for all $\,n.$  
Therefore, more generally $\ k\ge 3\,\Rightarrow\,\gcd(f_n,f_{n+1},\ldots,f_{n+k})= 6 $