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 $b_{1}=3, b_{k+1}=a_{k}+b_{k}(k=1,2, \cdots)$. Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
$$
-、 \because S_{n}=2 a_{n}-1, a_{1}=S_{1}=2 a_{1}-1, \therefore a_{1}=1
$$

Also, $a_{k}=S_{k}-S_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right)=2 a_{k}-2 a_{k-1}$,
$$
\therefore \quad a_{k}=2 a_{k-1} \text {. }
$$

Therefore, $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2.
For $b_{k+1}=a_{k}+b_{k}$, taking $k=1,2, \cdots, n-1$, we get
$$
\begin{array}{l}
b_{2}=a_{1}+b_{1}, \\
b_{3}=a_{2}+b_{2}, \\
\cdots \cdots \\
b_{n}=a_{n-1}+b_{n-1} .
\end{array}
$$

Adding both sides of the above equations, we get
$$
b_{n}=S_{n-1}+b_{1}=\frac{2^{n-1}-1}{2-1}+3=2^{n-1}+2 .
$$

Therefore, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$
$$
S^{\prime}{ }_{n}=1+2+2^{2}+\cdots+2^{n-1}+2 n=2^{n}+2 n-1 .
$$