The sequence $(x_n)$ is determined by the conditions:
$x_0=1992,x_n=-\frac{1992}{n} \cdot \sum_{k=0}^{n-1} x_k$ for $n \geq 1$.
Find $\sum_{n=0}^{1992} 2^nx_n$.
1. **Initial Conditions and Recurrence Relation:**
   The sequence \((x_n)\) is defined by:
   \[
   x_0 = 1992
   \]
   and for \(n \geq 1\),
   \[
   x_n = -\frac{1992}{n} \sum_{k=0}^{n-1} x_k
   \]

2. **Simplifying the Recurrence Relation:**
   We start by examining the recurrence relation. For \(n \geq 1\),
   \[
   x_n = -\frac{1992}{n} \sum_{k=0}^{n-1} x_k
   \]
   Let's denote \(S_n = \sum_{k=0}^{n} x_k\). Then,
   \[
   S_n = S_{n-1} + x_n
   \]
   Substituting \(x_n\) from the recurrence relation,
   \[
   S_n = S_{n-1} - \frac{1992}{n} S_{n-1}
   \]
   \[
   S_n = S_{n-1} \left(1 - \frac{1992}{n}\right)
   \]

3. **Finding a General Expression for \(x_n\):**
   We need to find a pattern or closed form for \(x_n\). Notice that:
   \[
   x_n = \frac{n}{n-1993} x_{n-1}
   \]
   This can be derived by manipulating the recurrence relation:
   \[
   x_n = -\frac{1992}{n} \sum_{k=0}^{n-1} x_k
   \]
   Using the fact that \(S_{n-1} = \sum_{k=0}^{n-1} x_k\),
   \[
   x_n = -\frac{1992}{n} S_{n-1}
   \]
   Since \(S_n = S_{n-1} + x_n\),
   \[
   S_n = S_{n-1} - \frac{1992}{n} S_{n-1}
   \]
   \[
   S_n = S_{n-1} \left(1 - \frac{1992}{n}\right)
   \]
   This implies:
   \[
   x_n = \frac{n}{n-1993} x_{n-1}
   \]

4. **Inductive Hypothesis:**
   Assume \(x_n = 1992(-1)^n \binom{1992}{n}\). We will prove this by induction.
   - **Base Case:**
     For \(n = 0\),
     \[
     x_0 = 1992
     \]
     which matches the given initial condition.
   - **Inductive Step:**
     Assume \(x_k = 1992(-1)^k \binom{1992}{k}\) holds for \(k = 0, 1, \ldots, n-1\). We need to show it holds for \(k = n\).
     \[
     x_n = \frac{n}{n-1993} x_{n-1}
     \]
     By the inductive hypothesis,
     \[
     x_{n-1} = 1992(-1)^{n-1} \binom{1992}{n-1}
     \]
     Therefore,
     \[
     x_n = \frac{n}{n-1993} \cdot 1992(-1)^{n-1} \binom{1992}{n-1}
     \]
     Using the identity \(\binom{1992}{n} = \frac{1992}{n} \binom{1991}{n-1}\),
     \[
     x_n = 1992(-1)^n \binom{1992}{n}
     \]
     This completes the induction.

5. **Summing the Series:**
   We need to find:
   \[
   \sum_{n=0}^{1992} 2^n x_n
   \]
   Using the binomial theorem,
   \[
   \sum_{n=0}^{1992} 2^n x_n = 1992 \sum_{n=0}^{1992} 2^n (-1)^{1992-n} \binom{1992}{n}
   \]
   By the binomial theorem,
   \[
   \sum_{n=0}^{1992} \binom{1992}{n} (-2)^n = (1 - 2)^{1992} = (-1)^{1992} = 1
   \]
   Therefore,
   \[
   \sum_{n=0}^{1992} 2^n x_n = 1992 \cdot 1 = 1992
   \]

The final answer is \(\boxed{1992}\).