1. Let $a$ be a natural number, then the units digit of $a+a^{2}+a^{3}+\cdots+$ $a^{2002}$ cannot be ( ).
(A) 0
(B) 2
(C) 4
(D) 6
-、1.C.
We only need to discuss $a$ being the ten digits $0,1,2, \cdots, 9$.
When $a=1$ or 6, the unit digit of $a+a^{2}+\cdots+a^{2000}$ is 2.

When $a \neq 1$ and $a \neq 6$, we have $a^{4 n}+a^{4 n+1}+a^{4 n+2}+$ $a^{4 n+3}$ (where $n$ is a positive integer) has a unit digit of 0. Therefore, the proposition is equivalent to finding the unit digit of $a+a^{2}$, which is the same as finding the unit digit of $a(a+1)$. Clearly, its unit digit cannot be 4.