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 $b_{1}=3, b_{k+1}=b_{k}+a_{k}(k=1,2, \cdots)$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
Given $S_{n}=2 a_{n}-1$ and $a_{1}=S_{1}=2 a_{1}-1$, we get $a_{1}=1$.
Also, $a_{n}=S_{n}-S_{n-1}=2 a_{n}-2 a_{n-1}$, which means $a_{n}=2 a_{n-1}$, thus $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2, so $a_{n}=2^{n-1}$.

Since $b_{k+1}=b_{k}+a_{k}$, taking $k=1,2, \cdots, n-1$ we get $b_{2}=a_{1}+b_{1}, b_{3}=a_{2}+b_{2}, \cdots, b_{n}=a_{n-1}+b_{n-1}$. Adding the above $n-1$ equations, we get $b_{n}=b_{1}+a_{1}+a_{2}+\cdots+a_{n}=3+\left(1+2+2^{2}+\cdots+2^{n-1}\right)=3+\frac{2^{n-1}-1}{2-1}=2^{n-1}+2$. Therefore, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is
$$
\begin{aligned}
{S^{\prime}}^{\prime} & =(2+1)+(2+2)+\left(2+2^{2}\right)+\cdots+\left(2+2^{n-1}\right) \\
& =2 n+2^{n}-1
\end{aligned}
$$