A round table has room for n diners ( $n\ge  2$). There are napkins in three different colours. In how many ways can the napkins be placed, one for each seat, so that no two neighbours get napkins of the same colour?
1. **Define the problem and initial conditions:**
   We need to find the number of ways to place napkins of three different colors around a round table with \( n \) seats such that no two adjacent seats have napkins of the same color. Let \( f(n) \) denote the number of such satisfactory placements for \( n \ge 2 \).

2. **Base case:**
   For \( n = 2 \), there are 3 choices for the first napkin and 2 choices for the second napkin (since it must be different from the first). Therefore, \( f(2) = 3 \times 2 = 6 \).

3. **Recurrence relation:**
   For \( n \ge 3 \), consider the following:
   - There are 3 choices for the first napkin.
   - For each subsequent napkin, there are 2 choices (since it must be different from the previous one).
   - This gives \( 3 \times 2^{n-1} \) ways to place the napkins if we ignore the circular nature of the table.
   - However, this count includes cases where the last napkin is the same color as the first napkin, which we need to subtract.

   Therefore, the recurrence relation is:
   \[
   f(n) = 3 \cdot 2^{n-1} - f(n-1)
   \]

4. **Finding a pattern:**
   By experimenting with small values of \( n \), we observe that \( f(n) \) can be expressed in terms of Jacobsthal numbers \( J_k \). The Jacobsthal numbers are defined by the recurrence relation:
   \[
   J_k = J_{k-1} + 2J_{k-2}
   \]
   with initial conditions \( J_0 = 0 \) and \( J_1 = 1 \).

5. **Expressing \( f(n) \) in terms of Jacobsthal numbers:**
   We find that \( f(n) = 6 \cdot J_{n-1} \) for \( n \ge 2 \).

6. **Jacobsthal numbers formula:**
   The closed-form expression for the \( k \)-th Jacobsthal number is:
   \[
   J_k = \frac{2^k - (-1)^k}{3}
   \]

7. **Simplifying \( f(n) \):**
   Substituting the closed-form expression of \( J_{n-1} \) into \( f(n) \):
   \[
   f(n) = 6 \cdot J_{n-1} = 6 \cdot \frac{2^{n-1} - (-1)^{n-1}}{3} = 2^n + 2(-1)^n
   \]

8. **Conclusion:**
   The number of ways to place the napkins such that no two adjacent seats have the same color is given by:
   \[
   f(n) = 2^n + 2(-1)^n
   \]

The final answer is \( \boxed{ f(n) = 2^n + 2(-1)^n } \).