Example 6 The adjacent terms $a_{n}, a_{n-1}$ of the sequence $\left\{a_{n}\right\}$ are the two roots of the equation $x^{2}-c_{n} x+\left(\frac{1}{3}\right)^{n}=0$, and $a_{1}=2$. Find the sum of the first $2 n$ terms of the sequence $\left\{c_{n}\right\}$, $S_{2 n}$.
Solution: According to the problem, we have
\[
\left\{
\begin{array}{l}
c_{n}=a_{n}+a_{n+1} \cdots \text { (1) } \\
a_{n} a_{n+1}=\left(\frac{1}{3}\right)^{n} \cdots(2)
\end{array}
\right.
\]
From (2), we get \(a_{n+1} a_{n+2}=\left(\frac{1}{3}\right)^{n+1} \cdots\) (3)

Therefore, \(a_{1}, a_{3}, \cdots, a_{2 k-1}, \cdots\) and \(a_{2}, a_{4}, \cdots, a_{2 k}, \cdots\) are both geometric sequences with a common ratio of \(\frac{1}{3}\).
From \(a_{1} a_{2}=\frac{1}{3}\) and \(a_{1}=2\), we get \(a_{2}=\frac{1}{6}\).
From \(a_{2} a_{3}=\frac{1}{9}\), we get \(a_{3}=\frac{2}{3}\).
From (1), we have
\[
\begin{aligned}
S_{2 n} & =c_{1}+c_{2}+\cdots+c_{2 n}=\left(a_{1}+a_{2}+\cdots+a_{2 n}\right)+\left(a_{2}+a_{3}+\cdots+a_{2 n+1}\right) \\
& =2\left(a_{2}+a_{4}+\cdots+a_{2 n}\right)+\left(a_{1}+a_{3}+\cdots+a_{2 n-1}\right)+\left(a_{3}+a_{5}+\cdots+a_{2 n+1}\right) \\
& =2 \cdot \frac{\frac{1}{6}\left[1-\left(\frac{1}{3}\right)^{n}\right]}{1-\frac{1}{3}}+\frac{2\left[1-\left(\frac{1}{3}\right)^{n}\right]}{1-\frac{1}{3}}+\frac{\frac{2}{3}\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{1}{3}} \\
& =\frac{3}{2}\left[1-\left(\frac{1}{3}\right)^{n}\right]\left(2 \times \frac{1}{6}+2+\frac{2}{3}\right) \\
& =\frac{9}{2}\left[1-\left(\frac{1}{3}\right)^{n}\right] .
\end{aligned}
\]

Analysis: Here, \(S_{2 n-1}\) can also be found using a similar method, but the result is not as neat as \(S_{2 n}\), so here we only require \(S_{2 n}\).