Example 5 Let $a_{1}=1, a_{2}=\frac{5}{3}, a_{n+2}=\frac{5}{3} a_{n+1}-\frac{2}{3} a_{n},(n=1,2, \cdots)$.
(1) Let $b_{n}=a_{n+1}-a_{n}(n=1,2, \cdots)$, find the general term formula of the sequence $\left\{b_{n}\right\}$;
(2) Find the sum of the first $n$ terms $S_{n}$ of the sequence $\left\{n a_{n}\right\}$.
Analyze: Use the known conditions to find the relationship between $b_{n}$ and $b_{n+1}$, and then use the method of difference for the product of an arithmetic sequence and a geometric sequence to solve such problems.

Solution (1) Since $b_{n+1}=a_{n+2}-a_{n+1}=\frac{5}{3} a_{n+1}-\frac{2}{3} a_{n}-a_{n+1}=\frac{2}{3}\left(a_{n+1}-a_{n}\right)=$ $\frac{2}{3} b_{n}$, hence $\left\{b_{n}\right\}$ is a geometric sequence with a common ratio of $\frac{2}{3}$, and $b_{1}=a_{2}-a_{1}=\frac{2}{3}$, so $b_{n}=\left(\frac{2}{3}\right)^{n}(n=1$, $2, \cdots)$.
(2) From $b_{n}=a_{n+1}-a_{n}=\left(\frac{2}{3}\right)^{n}$, we get $a_{n+1}-a_{1}=\left(a_{n+1}-a_{n}\right)+\left(a_{n}-a_{n-1}\right)+\cdots+$ $\left(a_{2}-a_{1}\right)=\left(\frac{2}{3}\right)^{n}+\left(\frac{2}{3}\right)^{n-1}+\cdots+\left(\frac{2}{3}\right)^{2}+\frac{2}{3}=2\left[1-\left(\frac{2}{3}\right)^{n}\right]$, noting that $a_{1}=1$, we can get $a_{n}=3-\frac{2^{n}}{3^{n-1}}(n=1,2, \cdots)$. Let the sum of the first $n$ terms of the sequence $\left\{\frac{n 2^{n-1}}{3^{n-1}}\right\}$ be $T_{n}$, then $T_{n}=1+$ $2 \cdot \frac{2}{3}+\cdots+n \cdot\left(\frac{2}{3}\right)^{n-1}$ (1), $\frac{2}{3} T_{n}=\frac{2}{3}+2 \cdot\left(\frac{2}{3}\right)^{2}+\cdots+n \cdot\left(\frac{2}{3}\right)^{n}(2)$. Subtracting the two equations gives $\frac{1}{3} T_{n}=1+\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\cdots+\left(\frac{2}{3}\right)^{n-1}-n\left(\frac{2}{3}\right)^{n}=3\left[1-\left(\frac{2}{3}\right)^{n}\right]-$ $n\left(\frac{2}{3}\right)^{n}$, hence $T_{n}=9\left[1-\left(\frac{2}{3}\right)^{n}\right]-3 n\left(\frac{2}{3}\right)^{n}=9-\frac{(3+n) 2^{n}}{3^{n-1}}$. Therefore, $S_{n}=a_{1}+$ $2 a_{2}+\cdots+n a_{n}=3(1+2+\cdots+n)-2 T_{n}=\frac{3}{2} n(n+1)+\frac{(3+n) 2^{n+1}}{3^{n-1}}-18$