If I let $n \geq 1$ be an integer and use a $2 \times n$ board $D_n$
containing $2n$ cells, each side has a length of 1. T
The brick can be vetical or horizontal containg $2$ cells(explained in the bottom part of the above picture).
The tiling of board ($D_n$) is placed with bricks so that :
1.item the bricks exactly cover $D_n$ and
2.item no two bricks overlap.
So for $n \geq 1$, we let $x_n$ become number of different tilings of the board $D_n$. Now I have to determine the value of $a_n$
I know there are only two ways to cover the left-most two squares: either a single vertical brick, or two horizontal bricks and that if you cover those two left ones with a single vertical brick, after that you have to choose any covering of the other $n-1$ columns and the only way to do this is precisely $x_{n-1}$. So, there are $x_{n-1}$ arrangements in which you cover the left two by a vertical brick and if you need two horizontal bricks, Then you've knocked out two columns, and need to pick a covering of the remaining $n-2$ columns. So now what would $x_n$ equal I think $a_n = a_(n-1) + a_(n-2)$ I am I right?
You are correct. Good thought