1. Let $\left\{a_{n}\right\}_{n \geqslant 1}$ be an arithmetic sequence with the first term 2 and common difference 1, and $\left\{b_{n}\right\}_{n \geqslant 1}$ be a geometric sequence with the first term 1 and common ratio 2. Let
$$
M_{n}=a_{b_{1}}+a_{b_{2}}+\cdots+a_{b_{n}}(n=1,2, \cdots) \text {. }
$$

Then the number of terms in $\left\{M_{n}\right\}_{n \geqslant 1}$ that do not exceed 2008 is $(\quad)$.
(A) 8
(B) 9
(C) 10
(D) 11
-、1.C.
Notice
$$
\begin{array}{l}
M_{n}=a_{b_{1}}+a_{b_{2}}+\cdots+a_{b_{n}} \\
=\left[2+\left(b_{1}-1\right)\right]+\cdots+\left[2+\left(b_{n}-1\right)\right] \\
=n+\left(1+2^{1}+2^{2}+\cdots+2^{n-1}\right) \\
=n+2^{n}-1 .
\end{array}
$$

From $n+2^{n}-1 \leqslant 2008$, we get $n \leqslant 10$. Therefore, the maximum value of $n$ is 10.