3. The sum of the first $n$ terms of the sequence $\left\{a_{n}!\right.$ is $S_{n}$, given that $a_{1}=4, a_{n}=S_{n-1}+2 n+1$, then $a_{2003}=(\quad)$.
A. $3 \times 2^{2003}-2$
B. $6 \times 2^{2003}-2$
C. $11 \times 2^{21002}-2$
D. $11 \times 2^{2001}-2$
3. D

From $a_{n}=S_{n-1}+2 n+1$, we get $a_{n+1}=S_{n}+2 n+3$. Subtracting the two equations, we get $a_{n+1}=2 a_{n}+2(n \geq 2)$, which means $a_{n+1}+2=2\left(a_{n}+2\right)$. Therefore, the sequence $\left\{a_{n}+2 \mid\right.$ starting from the 2nd term is a geometric sequence.