2. If the sequence satisfies $a_{1}=2, a_{n+1}=a_{n}+2 n$, then $a_{100}$ equals ( ).
A. 9900
B. 9902
C. 9904
D. 10100
2. B.

From the conditions, we have $a_{n+1}-a_{n}=2 n,(n \geqslant 1), a_{n}-a_{n-1}=2(n-1), \cdots, a_{2}-a_{1}=2$. Adding them up, we get $a_{n}-a_{1}=2(1+2+\cdots+n-1)=n(n-1)$, so $a_{n}=n^{2}-n+2$. Therefore, $a_{100}=9902$.