Example 3 A container holds 10 kilograms of water. The first time, $\frac{1}{3}$ is poured out, and then 1 kilogram of pure alcohol is added; the second time, $\frac{1}{3}$ is poured out again, and then $\frac{1}{2}$ kilogram of pure alcohol is added; thereafter, each time $\frac{1}{3}$ is poured out, and the amount of pure alcohol added is half of the previous addition. Ask: After the $n$-th operation, what is the concentration of the alcohol solution in the container?
Solution: Let the amount of water in the container after the $n$-th operation be $x_{n}$ kilograms, and the amount of pure alcohol be $a_{n}$ kilograms. Clearly, we have
$$
\begin{array}{c}
x_{1}=10 \times \frac{2}{3}=\frac{20}{3}, \\
x_{n}=\frac{2}{3} x_{n-1},
\end{array}
$$

Thus, $x_{n}=\frac{20}{3}\left(\frac{2}{3}\right)^{n-1}$.
Since $a_{1}=1$,
According to the problem, we have $a_{n}=\frac{2}{3} a_{n-1}+\left(\frac{1}{2}\right)^{n-1}(n \geqslant 2)$.
Let
$$
a_{n}+m\left(\frac{1}{2}\right)^{n}=\frac{2}{3}\left[a_{n-1}+m\left(\frac{1}{2}\right)^{n-1}\right],
$$

Thus,
$$
a_{n}=\frac{2}{3} a_{n-1}+\frac{m}{6}\left(\frac{1}{2}\right)^{n-1} \text {, }
$$

Comparing the coefficients of (1) and (2), we get $\quad \frac{m}{6}=1, m=6$,
That is,
$$
a_{n}+6 \cdot\left(\frac{1}{2}\right)^{n}=\frac{2}{3}\left[a_{n-1}+6 \cdot\left(\frac{1}{2}\right)^{n-1}\right] \text {, }
$$

Thus,
$$
a_{n}=4 \cdot\left(\frac{2}{3}\right)^{n-1}-6 \cdot\left(\frac{1}{2}\right)^{n} \text {. }
$$

Therefore, the alcohol concentration in the container after the $n$-th operation is
$$
\frac{a_{n}}{x_{n}+a_{n}}=\frac{12-9\left(\frac{3}{4}\right)^{n-1}}{32-9\left(\frac{3}{4}\right)^{n-1}} \text {, }
$$

Answer: The alcohol concentration in the container after the $n$-th operation is $\frac{12-9\left(\frac{3}{4}\right)^{n-1}}{32-9\left(\frac{3}{4}\right)^{n-1}}$.