6. (10 points) The calculation result of the expression $1!\times 3-2!\times 4+3!\times 5-4!\times 6+\cdots+2009!\times 2011-2010!\times 2012+2011$ ! is $\qquad$
【Solution】Solution: Grouping to find the pattern:
$$
\begin{array}{l}
2009! \times 2011 - 2010! \times 2012 + 2011! \\
= 2009!(2011 - 2010 \times 2012 + 2010 \times 2011) \\
= 2009!
\end{array}
$$

Then for $2007! \times 2009 - 2008! \times 2010 + 2009!$
$$
\begin{array}{l}
= 2007!(2009 - 2008 \times 2010 + 2008 \times 2009) \\
= 2007!
\end{array}
$$

By the odd term forward split cancellation rule we get
$$
\begin{array}{l}
\text { Original expression } = 2009! \times 2011 - 2010! \times 2012 + 2011! + \cdots + 5! \times 7 - 4! \times 6 + 3! \times 5 - 2! \times 4 + 1! \times 3 \\
= 1! \\
= 1
\end{array}
$$

Therefore, the answer is: 1