9. (This sub-question is worth 15 points) Let $a_{n}=2^{n}, n \in \mathbf{N}^{+}$, and the sequence $\left\{b_{n}\right\}$ satisfies $b_{1} a_{n}+b_{2} a_{n-1}+\cdots+b_{n} a_{1}=2^{n}-\frac{n}{2}-1$. Find the sum of the first $n$ terms of the sequence $\left\{a_{n} \cdot b_{n}\right\}$.
From $a_{1}=2$ and $b_{1} a_{1}=2-\frac{1}{2}-1$, we can get $b_{1}=\frac{1}{4}$.
5 points

When $n \geq 2$, from the given conditions, we have $\left\{\begin{array}{ll}b_{1} \cdot 2^{n-1}+b_{2} \cdot 2^{n-2}+\cdots+b_{n-1} \cdot 2=2^{n-1}-\frac{n-1}{2}-1 & , \cdots \cdots \text { (1) } \\ b_{1} \cdot 2^{n}+b_{2} \cdot 2^{n-1}+\cdots+b_{n-1} \cdot 2^{2}+b_{n} \cdot 2=2^{n}-\frac{n}{2}-1 & , \cdots \cdots \text { (2) }\end{array}\right.$
Multiplying both sides of (1) by 2, we get $b_{1} \cdot 2^{n}+b_{2} \cdot 2^{n-1}+\cdots+b_{n-1} \cdot 2^{2}=2^{n}-n-1$.
10 points

From (2) and (3), we can find that $b_{n}=\frac{n}{4}$.
10 points
Thus, $a_{n} \cdot b_{n}=\frac{n}{4} \cdot 2^{n}$. Let $T_{n}=\sum_{k=1}^{n} a_{k} \cdot b_{k}$, then we have $\left\{\begin{array}{l}T_{n}=\frac{1}{4} \sum_{k=1}^{n} k \cdot 2^{k} \\ 2 T_{n}=\frac{1}{4} \sum_{k=1}^{n} k \cdot 2^{k+1}\end{array}\right.$, therefore
$$
T_{n}=\frac{1}{4}\left(n \cdot 2^{n+1}-\sum_{k=1}^{n} 2^{k}\right)=\frac{(n-1) \cdot 2^{n}+1}{2}
$$
15 points