34. Suppose $x_{0}, x_{1}, x_{2}, \cdots$ is a sequence of numbers such that $x_{0}=1000$, and
$$
x_{n}=-\frac{1000}{n}\left(x_{0}+x_{1}+x_{2}+\cdots+x_{n-1}\right)
$$
for all $n \geq 1$. Find the value of
$$
\frac{1}{2^{2}} x_{0}+\frac{1}{2} x_{1}+x_{2}+2 x_{3}+2^{2} x_{4}+\cdots+2^{997} x_{999}+2^{998} x_{1000}-
$$
34. Ans: 250
Observe that when $n \geq 2$,
$$
\left\{\begin{array}{cl}
n x_{n} & =-1000\left(x_{0}+x_{1}+\cdots+x_{n-1}\right), \\
(n-1) x_{n-1} & =-1000\left(x_{0}+x_{1}+\cdots+x_{n-2}\right) .
\end{array}\right.
$$

Thus, we have
$$
n x_{n}-(n-1) x_{n-1}=-1000 x_{n-1} \Longrightarrow x_{n}=-\left(\frac{1000-(n-1)}{n}\right) x_{n-1}
$$

It is easy to check that the above formula holds even when $n=1$. Therefore, for $1 \leq n \leq 100$, we have
$$
\begin{aligned}
x_{n} & =-\frac{1000-(n-1)}{n} x_{n-1} \\
& =(-1)^{2} \frac{1000-(n-1)}{n} \cdot \frac{1000-(n-2)}{n-1} x_{n-2} \\
& =\cdots \\
& =(-1)^{n} \frac{1000-(n-1)}{n} \cdot \frac{1000-(n-2)}{n-1} \cdots \frac{1000}{1} x_{0} \\
& =(-1)^{n}\binom{1000}{n} x_{0} .
\end{aligned}
$$

Hence we have
$$
\begin{aligned}
& \frac{1}{2^{2}} x_{0}+\frac{1}{2} x_{1}+x_{2}+\cdots+2^{998} x_{1000} \\
= & \frac{1}{4}\left(x_{0}+2 x_{1}+2^{2} x_{2}+\cdots+2^{1000} x_{1000}\right) \\
= & \frac{1}{4}\left(x_{0}-\binom{1000}{1} 2 x_{0}+\binom{1000}{2} 2^{2} x_{0}-\binom{1000}{3} 2^{3} x_{0}+\cdots+\binom{1000}{1000} 2^{1000} x_{0}\right) \\
= & \frac{x_{0}}{4}\left(1-\binom{1000}{1} 2+\binom{1000}{2} 2^{2}-\binom{1000}{2} 2^{3}+\cdots+\binom{1000}{1000} 2^{1000}\right) \\
= & \frac{1000}{4}(1-2)^{1000} \\
= & 250 .
\end{aligned}
$$