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 $b_{1}=3$, $b_{k+1}=a_{k}+b_{k}(k=1,2, \cdots)$. Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
Given $S_{n}=2 a_{n}-1, n \geqslant 2$, then $S_{n-1}=2 a_{n-1}-1$.
Subtracting the two equations gives $a_{n}=2\left(a_{n}-a_{n-1}\right) \Rightarrow a_{n}=2 a_{n-1} \Rightarrow a_{n}=2^{n-1}(n=1,2, \cdots)$.
Thus, $b_{n}=\sum_{k=1}^{n-1}\left(b_{k+1}-b_{k}\right)+b_{1}=\sum_{k=1}^{n-1} a_{k}+b_{1}=\sum_{k=1}^{n-1} 2^{k-1}+3=2^{n-1}+2$.
Therefore, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $T_{n}=\sum_{k=1}^{n} b_{k}=\sum_{k=1}^{n} 2^{k-1}+2 n=2^{n}+2 n-1$.