2. For a finite sequence $P=\left(p_{1}, p_{2}, \cdots\right.$, $p_{n}$ ), the Cesàro sum of $P$ (Cesàro being a mathematician's name) is defined as $\frac{s_{1}+s_{2}+\cdots+s_{n}}{n}$, where $s_{k}=p_{1}+p_{2}+$ $\cdots+p_{k}(1 \leqslant k \leqslant n)$. If a 99-term sequence $\left(p_{1}\right.$, $p_{2}, \cdots, p_{99}$ ) has a Cesàro sum of 1000, then the Cesàro sum of the 100-term sequence $\left(1, p_{1}, p_{2}, \cdots, p_{99}\right)$ is ( ).
(A) 990
(B) 991
(C) 992
(D) 993
2. (B).
$$
\begin{array}{l}
\because \frac{s_{1}+s_{2}+\cdots+s_{99}}{99}=1000, \\
\therefore s_{1}+s_{2}+\cdots+s_{99}=99000 .
\end{array}
$$

Therefore, $\frac{1+\left(1+s_{1}\right)+\left(1+s_{2}\right)+\cdots+\left(1+s_{99}\right)}{100}$
$$
\begin{array}{l}
=\frac{100+\left(s_{1}+s_{2}+\cdots+s_{99}\right)}{100} \\
=\frac{100+99000}{100}=991 .
\end{array}
$$