1. If the coefficient of $x$ in the expansion of $(1+x)(1+3 x) \cdots(1+2001 x)$ is $n$, then $\mathrm{i}^{n}$ equals ( ).
(A)i
(B) $-\mathrm{i}$
(C) 1
(D) -1
- 1. (A).
$(1+x)(1+3x) \cdots(1+2001x)$ The coefficient of the linear term in the expanded form is $1+3+\cdots+2001=\frac{1}{2} \times 2002 \times 1001=$ $1001^{2}$.
Therefore, $i^{n}=i^{1001^{2}}=i$.