9. Given $f(x)=2^{x}$, then $f(1) \cdot f^{-1}(2)+f(2) \cdot f^{-1}(4)+\cdots+f(n) \cdot f^{-1}$ $\left(2^{n}\right)=$
9. $(n-1) 2^{n+1}+2$

Since $a_{k}=f(k) \cdot f^{-1}\left(2^{k}\right)=k \cdot 2^{k}$, we have $k \cdot 2^{k}=(k+1) \cdot 2^{k+1}-k \cdot 2^{k}-2^{k+1}$, so
$$
\begin{aligned}
\sum_{k=1}^{n} k \cdot 2^{k} & =\sum_{k=1}^{n}(k+1) 2^{k+1}-\sum_{k=1}^{n} k \cdot 2^{k}-\sum_{k=1}^{n} 2^{k+1} \\
& =\sum_{k=1}^{n+1} k \cdot 2^{k}-\sum_{k=1}^{n} k \cdot 2^{k}-\frac{4\left(1-2^{n}\right)}{1-2} \\
& =(n-1) \cdot 2^{n+1}+2
\end{aligned}
$$