19. The sequence of real numbers $x_{0}, x_{1}, x_{2}, \cdots$, is defined as $x_{0}=1989, x_{n}=-\frac{1989}{n}$ $\sum_{k=0}^{n-1} x_{k}(n \geqslant 1)$, calculate the value of $\sum_{n=0}^{1988} 2^{n} x_{n}$.
19. Let $n=1989$, when $n \geqslant 2$, we have $n x_{n}=-m \sum_{k=0}^{n-1} x_{k}$ and $(n-1) x_{n-1}=$ $-m \sum_{k=0}^{n-2} x_{k}$. Subtracting the two equations gives $x_{n}=-\frac{m-n+1}{n} x_{n-1}(n \geqslant 2)$. This formula also holds when $n=1$, so for $1 \leqslant n \leqslant m$, we have $x_{n}=-\frac{(m-n+1)}{n} x_{n-1}==\cdots=(-1)^{n} C_{m}^{n} \cdot m$.
Using the binomial theorem, we get
$$
\sum_{n=0}^{1989} 2^{n} x_{n}=m \sum_{n=0}^{m} 2^{n}(-1)^{n} C_{m}^{n}=m(1-2)^{m}=-m=-1989 .
$$