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 $\left\{b_{n}\right\}$.
$$
2^{n}+2 n-1
$$
- [Analysis and Solution]

From $S_{n}=2 a_{n}-1$, we know $a_{1}=S_{1}=2 a_{1}-1$, which gives $a_{1}=1$.
And $a_{k}=S_{k}-S_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right)=2 a_{k}-2 a_{k-1}$.
So $a_{k}=2 a_{k-1}$, thus $a_{k}=2^{k-1}$. ( $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2 )
Given $b_{k+1}=a_{k}+b_{k}$, we have $b_{n}=b_{1}+a_{1}+a_{2}+\cdots+a_{n-1}=3+2^{n-1}-1=2^{n-1}+2$, so the sum of the first $n$ terms of the sequence is:
$$
S_{n}^{\prime}=1+2+2^{2}+\cdots+2^{n-1}+2 n=2^{n}+2 n-1 \text {. }
$$