1) The sequence $\{2 n-1\}\left(n \in \mathbf{N}^{*}\right)$ is grouped in the original order with the requirement that the $n$-th group has $2^{n}$ terms. Then 2009 is in the ( ) group.
(A) 7
(B) 8
(C) 9
(D) 10
(1) $\mathrm{C}$ Hint: The sum of the number of terms in the first $n$ groups is $2+2^{2}+\cdots+2^{n}=2^{n+1}-2$,
2009 is the 1005th term of the sequence.
$$
510=2^{9}-2<1005<2^{10}-2=1022 .
$$

Therefore, 2009 is in the 9th group, choose C.