2. Let $n=9+99+\cdots+\underbrace{99 \cdots 9}_{99 \uparrow}$. Then in the decimal representation of $n$, the digit 1 appears ( ) times.
(A) 50
(B) 90
(C) 99
(D) 100
2.C.

Since $n=(10-1)+(100-1)+\cdots+(1 \underbrace{00 \cdots 0}_{99 \uparrow}-1)$ $=\underbrace{11 \cdots 10}_{99}-99=\underbrace{11 \cdots 1}_{91 \uparrow} 011$,
therefore, in the decimal representation of $n$, the digit 1 appears $97+2=99$ (times).