The first letter can be anything. So there are 3 choices.
The second letter cannot be the same as the first letter. So there are 2 choices.
For the third letter, there are 2 choices. Same as the second letter.
We can continue this pattern until the end of the string.
Thus, there are $3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ valid strings.
Rewriting this, we get $V_n = 3 \cdot 2^{n-1}$.