2. 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 $\left\{b_{n}\right\}$.
(1996 National High School Competition Question)
2. From $S_{n}=2 a_{n}-1$, we know that $a_{1}=2 a_{1}-1$, which gives $a_{1}=1$. Also, $a_{k}=S_{k}-S_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right)=$ $2 a_{k}-2 a_{k-1}$, so $a_{k}=2 a_{k-1}$. Therefore, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2. From $b_{k+1}=a_{k}+b_{k}$, we get $b_{2}=a_{1}+b_{1}, b_{3}=a_{2}+b_{2}, \cdots, b_{n}=a_{n-1}+b_{n-1}$. Adding these equations, $b_{n}=S_{n-1}+b_{1}=\frac{2^{n-1}}{2-1}+$ $3=2^{n-1}+2$, so the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $S_{n}^{\prime}=1+2+\cdots+2^{n-1}+2 n=2^{n}+2 n-1$.