3.1.17 ** 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 infinite sequence $c_{1}, c_{2}, \cdots, c_{n}, \cdots$.
Given the problem, we have
\[
\left\{
\begin{array}{l}
c_{n}=a_{n}+a_{n+1}, \\
a_{n} a_{n+1}=\left(\frac{1}{3}\right)^{n}.
\end{array}
\right.
\]
(1)
From (2), we get \(a_{n+1} a_{n+2} - \left(\frac{1}{3}\right)^{n+1}\), so \(\frac{a_{n+2}}{a_{n}} = \frac{1}{3}\). Therefore,
\[
\begin{aligned}
\sum_{n=1}^{\infty} c_{n} & = \sum_{n=1}^{\infty} a_{n} + \sum_{n=2}^{\infty} a_{n} = \sum_{k=0}^{\infty} a_{2 k+1} + 2 \cdot \sum_{k=1}^{\infty} a_{2 k} + \sum_{k=1}^{\infty} a_{2 k+1} \\
& = 2 \cdot \sum_{k=0}^{\infty} a_{2 k+1} + 2 \cdot \sum_{k=1}^{\infty} a_{2 k} - a_{1}.
\end{aligned}
\]

Given \(a_{1} = 2\) and from (2), we get \(a_{2} = \frac{1}{6}\), so
\[
\sum_{n=1}^{\infty} c_{n} = 2 \times \frac{2}{1 - \frac{1}{3}} + 2 \times \frac{\frac{1}{6}}{1 - \frac{1}{3}} - 2 = \frac{9}{2}.
\]