13. 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 $b_{1}=3, b_{k+1}=a_{k}+b_{k}$ $(k=1,2,3, \cdots)$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
13. Since $S_{n}=2 a_{n}-1=2\left(S_{n}-S_{n-1}\right)-1 \Rightarrow S_{n}+1=2\left(S_{n-1}+1\right)(n \geqslant 2)$

Therefore, $S_{n}+1=\left(S_{1}+1\right) 2^{n-1}$, and since $S_{1}=a_{1}$, we have $a_{1}=2 a_{1}-1 \Rightarrow a_{1}=1$.
So, $S_{n}=2^{n}-1$. Let the sum of the first $n$ terms of $\left\{b_{n}\right\}$ be $S^{\prime}{ }_{n}$.
Thus, from $b_{k+1}=a_{k}+b_{k}$, we get $b_{n}=b_{n-1}+a_{n-1}, b_{n-1}=b_{n-2}+a_{n-2}, \cdots b_{2}=b_{1}+a_{1}$. Adding these up: $b_{n}=$ $b_{1}+\left(a_{1}+\cdots+a_{n-1}\right)=b_{1}+S_{n-1}=b_{1}+2^{n-1}-1$, and since $b_{1}=3$, we have $b_{n}=2+2^{n-1}$, so $S_{n}^{\prime}=2 n+\sum_{k=1}^{n} 2^{k-1}$ $=2 n+\frac{1-2^{n}}{1-2}=2^{n}+2 n-1$.