1. For a finite sequence $P=\left(p_{1}, p_{2}, \cdots, p_{n}\right)$, 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 $\left(p_{1}, p_{2}, \cdots, p_{99}\right)$ 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
1. B Because $\frac{s_{1}+s_{2}+\cdots+s_{99}}{99}=1000$, so $s_{1}+s_{2}+\cdots+s_{99}=99000$.

Thus $\frac{1+\left(1+s_{1}\right)+\left(1+s_{2}\right)+\cdots+\left(1+s_{99}\right)}{100}=\frac{100+\left(s_{1}+s_{2}+\cdots+s_{99}\right)}{100}=\frac{100+99000}{100}=991$. Therefore, the answer is B.