The sum
$$\sum_{n=0}^{2016\cdot2017^2}2018^n$$
can be represented uniquely in the form $\sum_{i=0}^{\infty}a_i\cdot2017^i$ for nonnegative integers $a_i$ less than $2017$. Compute $a_0+a_1$.
To solve the problem, we need to represent the sum \( \sum_{n=0}^{2016 \cdot 2017^2} 2018^n \) in the form \( \sum_{i=0}^{\infty} a_i \cdot 2017^i \) where \( 0 \leq a_i < 2017 \). We are asked to compute \( a_0 + a_1 \).

1. **Understanding the Sum**:
   The given sum is a geometric series with the first term \( a = 1 \) and common ratio \( r = 2018 \). The number of terms in the series is \( 2016 \cdot 2017^2 + 1 \).

2. **Sum of the Geometric Series**:
   The sum of a geometric series \( \sum_{n=0}^{N} r^n \) is given by:
   \[
   S = \frac{r^{N+1} - 1}{r - 1}
   \]
   Here, \( r = 2018 \) and \( N = 2016 \cdot 2017^2 \). Therefore, the sum \( S \) is:
   \[
   S = \frac{2018^{2016 \cdot 2017^2 + 1} - 1}{2018 - 1} = \frac{2018^{2016 \cdot 2017^2 + 1} - 1}{2017}
   \]

3. **Modulo Representation**:
   We need to represent \( S \) in the form \( \sum_{i=0}^{\infty} a_i \cdot 2017^i \). This is equivalent to finding the base-2017 representation of \( S \).

4. **Simplifying the Expression**:
   Notice that \( 2018 \equiv 1 \pmod{2017} \). Therefore:
   \[
   2018^{2016 \cdot 2017^2 + 1} \equiv 1^{2016 \cdot 2017^2 + 1} \equiv 1 \pmod{2017}
   \]
   Thus:
   \[
   S = \frac{2018^{2016 \cdot 2017^2 + 1} - 1}{2017} \equiv \frac{1 - 1}{2017} \equiv 0 \pmod{2017}
   \]

5. **Finding \( a_0 \) and \( a_1 \)**:
   Since \( S \equiv 0 \pmod{2017} \), the coefficients \( a_0 \) and \( a_1 \) must satisfy:
   \[
   a_0 + a_1 \cdot 2017 \equiv 0 \pmod{2017}
   \]
   The simplest solution is \( a_0 = 0 \) and \( a_1 = 0 \).

6. **Conclusion**:
   Therefore, \( a_0 + a_1 = 0 + 0 = 0 \).

The final answer is \( \boxed{0} \).