$8 \cdot 36$ Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms
$$
S_{n}=2 a_{n}-1,(n=1,2, \cdots)
$$

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 $\left\{b_{n}\right\}$.
[Solution] From $s_{n}=2 a_{n}-1$, we know,
$$
\begin{array}{l}
a_{1}=2 a_{1}-1, \\
a_{1}=1 .
\end{array}
$$

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 $a_{k}=2 a_{k-1}$.
Thus, 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
$$
\left\{\begin{array}{l}
b_{2}=a_{1}+b_{1}, \\
b_{3}=a_{2}+b_{2}, \\
\cdots \cdots \cdots \cdots \\
b_{n}=a_{n-1}+b_{n-1} .
\end{array}\right.
$$

Adding these equations,
$$
b_{n}=s_{n-1}+b_{1}=\frac{2^{n-1}-1}{2-1}+3=2^{n-1}+2 .
$$

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