One. (This question is worth 25 points) 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 $\left\{b_{n}\right\}$.
$$
\begin{array}{l}
\text { -、 } \because S_{n}=2 a_{n}-1, a_{1}=S_{1}=2 a_{1}-1, \\
\begin{array}{l} 
\therefore a_{1}=1 . \\
\text { Also, } a_{k}=S_{k}-S_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right) \\
=2 a_{k}-2 a_{k-1},
\end{array}
\end{array}
$$
$\therefore a_{k}=2 a_{k-1}$. 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
$$
b_{n}=S_{n-1}+b_{1}=\frac{2^{n-1}-1}{2-1}+3=2^{n-1}+2 \text {. }
$$

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