$41 S_{n}$ represents the sum of the first $n$ terms of $\left\{a_{n}\right\}$, given that $a_{1}=4, a_{n}=S_{n-1}+2 n+1$, then $a_{1990}=（$ )
A. $3 \times 2^{1990}-2$
B. $6 \times 2^{1990}-2$
C. $11 \times 2^{1989}-2$
D. $11 \times 2^{1988}-2$
41 D.
$$
S_{n}=2 S_{n-1}+2 n+1, S_{n-1}=2 S_{n-2}+(2 n-1),
$$
\cdots, \( S_{2}=2 S_{1}+5 \). Multiply each of the above equations by \( 1, 2, \cdots, 2^{n-2} \), respectively, and then add them. After rearranging, we get
$$
S_{n}=(2 n+1)+2(2 n-1)+\cdots+2^{n-2} \times 5+2^{n-3} \times 8
$$

From the above, subtract
$$
S_{n-1}=(2 n-1)+2(2 n-3)+\cdots+2^{n-3} \times 5+2^{n-3} \times 8
$$

to get
$$
\begin{aligned}
a_{n} & =2+2 \times 2+\cdots+2 \times 2^{n-3}+2^{n-2} \times 5+2^{n-3} \times 8 \\
& =-2+2 \times 2^{n-2}+5 \times 2^{n-2}+4 \times 2^{n-2} \\
& =11 \times 2^{n-2}-2
\end{aligned}
$$