2. The sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms as $S_{n}$. It is known that $a_{1}=4, a_{n}=S_{n-1}+2 n+1$, then $a_{2005}$ equals
A. $3 \times 2^{2005}-2$
B. $6 \times 2^{2005}-2$
C. $11 \times 2^{2004}-2$
D. $11 \times 2^{2003}-2$
2. Choose D.

Brief solution: From $a_{n}=S_{n-1}+2 n+1$ we get $a_{n+1}=S_{n}+2 n+3$, subtracting the two equations yields $a_{n+1}=2 a_{n}+2(n \geqslant 2)$, which means $a_{n+1}+2=2\left(a_{n}+2\right)$. Thus, the sequence $\left\{a_{n}+2\right\}$ forms a geometric sequence from the 2nd term onwards, hence $a_{2005}=11 \times 2^{2003} - 2$.