11. (5 points) Arrange the odd numbers from 1 to 2011 in a sequence, then group them according to the pattern of 1, 2, 3, 2, 1, 2, 3, 2, 1, ... as follows (each parenthesis is a group):
(1) $(3,5)(7,9,11)(13,15)(17)(19,21)(23,25,27)(29,31)(33) \ldots$ The sum of the numbers in the last parenthesis is $\qquad$
【Solution】Solution: $1+2+3+2=8$, that is, the grouping rule is a cycle every 8 numbers,
$$
\begin{array}{l}
2010 \div 2+1=1006 \text { (numbers) } \\
1006 \div 8=125 \cdots 6
\end{array}
$$
In 2011, the last 6 odd numbers are: (2001), (2003, 2005), (2007, 2009, 2011). The sum of the numbers in the last parenthesis is: $2007+2009+2011=6027$. Therefore, the answer is: 6027.