1. There is a sequence of numbers: $1,1,5,13,25,41, \cdots$, arranged according to this pattern, the 101st number is ( ).
(A) 19801
(B) 19802
(C) 19901
(D) 19902
- 1. A.

Calculating the difference between two consecutive numbers, we know the $n$-th number is
$$
\begin{array}{l}
1+4 \times 1+4 \times 2+\cdots+4(n-2) \\
=1+4[1+2+\cdots+(n-2)] \\
=1+4 \times \frac{(n-1)(n-2)}{2} \\
=1+2(n-1)(n-2) .
\end{array}
$$

Therefore, the 101st number is
$$
1+2(101-1)(101-2)=19801 .
$$