Given an $m\times n$ grid with unit squares coloured either black or white, a black square in the grid is [i]stranded [/i]if there is some square to its left in the same row that is white and there is some square above it in the same column that is white.
Find a closed formula for the number of $2\times n$ grids with no stranded black square.
Note that $n$ is any natural number and the formula must be in terms of $n$ with no other variables.
1. **Define the problem and initial conditions:**
   Let \( N(n) \) denote the number of \( 2 \times n \) grids with no stranded black squares. We start by considering the base case:
   \[
   N(1) = 4
   \]
   This is because there are four possible colorings of a \( 2 \times 1 \) grid: both squares white, both squares black, top white and bottom black, and top black and bottom white.

2. **Recursive relation:**
   Consider a \( 2 \times n \) grid. We can think of it as a \( 2 \times (n-1) \) grid with an additional column added to the right. We need to count the valid colorings of this new column based on the colorings of the \( 2 \times (n-1) \) grid.

3. **Case 1: Bottom row of the \( 2 \times (n-1) \) grid is all black:**
   - There are \( 2^{n-1} \) such colorings.
   - For the rightmost column, there are no constraints, so there are 4 possible colorings.
   - Thus, the number of valid colorings in this case is:
     \[
     4 \cdot 2^{n-1}
     \]

4. **Case 2: Bottom row of the \( 2 \times (n-1) \) grid is not all black:**
   - There are \( N(n-1) - 2^{n-1} \) such colorings.
   - For the rightmost column, if the bottom-right square is black, the top-right square must be white to avoid a stranded black square. This gives 3 possible colorings.
   - Thus, the number of valid colorings in this case is:
     \[
     3 \cdot (N(n-1) - 2^{n-1})
     \]

5. **Combine the cases:**
   \[
   N(n) = 4 \cdot 2^{n-1} + 3 \cdot (N(n-1) - 2^{n-1})
   \]
   Simplify the expression:
   \[
   N(n) = 4 \cdot 2^{n-1} + 3N(n-1) - 3 \cdot 2^{n-1}
   \]
   \[
   N(n) = 3N(n-1) + 2^{n-1}
   \]

6. **Solve the recurrence relation:**
   We solve the recurrence relation \( N(n) = 3N(n-1) + 2^{n-1} \) with the initial condition \( N(1) = 4 \).

   Let's find a closed form by iterating the recurrence:
   \[
   N(2) = 3N(1) + 2^1 = 3 \cdot 4 + 2 = 12 + 2 = 14
   \]
   \[
   N(3) = 3N(2) + 2^2 = 3 \cdot 14 + 4 = 42 + 4 = 46
   \]
   We observe that the recurrence relation can be solved using the method of characteristic equations or by recognizing a pattern.

7. **General solution:**
   The homogeneous part of the recurrence \( N(n) = 3N(n-1) \) suggests a solution of the form \( N(n) = A \cdot 3^n \). The particular solution can be found by assuming a form \( B \cdot 2^n \).

   Combining these, we get:
   \[
   N(n) = A \cdot 3^n + B \cdot 2^n
   \]

8. **Determine constants \( A \) and \( B \):**
   Using the initial condition \( N(1) = 4 \):
   \[
   4 = A \cdot 3^1 + B \cdot 2^1
   \]
   \[
   4 = 3A + 2B
   \]

   Using the second condition \( N(2) = 14 \):
   \[
   14 = A \cdot 3^2 + B \cdot 2^2
   \]
   \[
   14 = 9A + 4B
   \]

   Solving these equations:
   \[
   3A + 2B = 4
   \]
   \[
   9A + 4B = 14
   \]

   Multiply the first equation by 2:
   \[
   6A + 4B = 8
   \]

   Subtract the first equation from the second:
   \[
   (9A + 4B) - (6A + 4B) = 14 - 8
   \]
   \[
   3A = 6
   \]
   \[
   A = 2
   \]

   Substitute \( A = 2 \) back into the first equation:
   \[
   3(2) + 2B = 4
   \]
   \[
   6 + 2B = 4
   \]
   \[
   2B = -2
   \]
   \[
   B = -1
   \]

   Thus, the closed form is:
   \[
   N(n) = 2 \cdot 3^n - 2^n
   \]

The final answer is \( \boxed{2 \cdot 3^n - 2^n} \).