For example, a $53 \times 2 n$ chessboard can be covered with $3 n$ rectangular pieces of size $1 \times 2$. Question: How many ways are there to completely cover the chessboard?
Let the number of covering methods that satisfy the conditions be $a_{n}$ (each two columns form a group, for a total of $n$ groups).

As shown in Figure 1, in a $3 \times 2n$ grid, when the first two columns are covered with three $1 \times 2$ rectangular pieces placed horizontally, the number of covering methods that satisfy the conditions is $b_{n}$; when the first column is covered with a $1 \times 2$ rectangular piece placed vertically, the number of covering methods that satisfy the conditions is $c_{n}$. Then,
$$
a_{n}=b_{n}+2 c_{n} \text {, and } b_{n}=a_{n-1} .
$$

As shown in Figure 1, when a $1 \times 2$ rectangular piece is placed vertically covering positions 1 and 2 in the first column, then positions 3 and 4 must be covered with a $1 \times 2$ rectangular piece placed horizontally, and there are two possibilities:
(1) A $1 \times 2$ rectangular piece is placed vertically covering positions 5 and 6;
(2) Two $1 \times 2$ rectangular pieces are placed horizontally covering positions 6 and 7, and 5 and 8, respectively.
$$
\begin{array}{l}
\text { Therefore, } c_{n}=a_{n-1}+c_{n-1} \text {. } \\
\text { Thus, } a_{n}=4 a_{n-1}-a_{n-2} \text {, and. } \\
a_{1}=3, a_{2}=3+2 \times 4=11 . \\
\text { Hence } a_{n}=\frac{\sqrt{3}+1}{2 \sqrt{3}}(2+\sqrt{3})^{n}+\frac{\sqrt{3}-1}{2 \sqrt{3}}(2-\sqrt{3})^{n} .
\end{array}
$$