Example 1 Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}=2 a_{n}-1(n=1,2, \cdots)$, and the sequence $\left\{b_{n}\right\}$ satisfies the conditions $b_{1}=3, b_{k+1}=a_{k}+b_{k}(k=1,2, \cdots)$. Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
From (1) we get $a_{1}=1$, and from $a_{n}=S_{n}-S_{n-1}=2 a_{n}-2 a_{n-1} \Rightarrow a_{n}=2 a_{n-1}$, hence $a_{n}=2^{n-1}, S_{n}=2^{n}-1$. Therefore, $b_{n}=\sum_{k=1}^{n}\left(b_{k+1}-b_{k}\right)+b_{1}=\sum_{k=1}^{n} a_{k}+b_{1}=3+S_{n}=$ $2+2^{n-1}$, thus $S_{n}=b_{1}+\cdots+b_{n}=2 n+\sum_{k=1}^{n-1} 2^{k}=2^{n}+2 n-1$.