(7) Let $f(x)=a x+b$, where $a, b$ are real numbers, $f_{1}(x)=f(x), f_{n+1}(x)=$ $f\left(f_{n}(x)\right), n=1,2,3, \cdots$, if $f_{7}(x)=128 x+381$, then $a+b=$ $\qquad$ .
(7) From the problem, we know
$$
\begin{aligned}
f_{n}(x) & =a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
& =a^{n} x+\frac{a^{n}-1}{a-1} \cdot b,
\end{aligned}
$$

From
$$
f_{7}(x)=128 x+381
$$

we get $a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381$, so $a=2, b=3$, thus $a+b=5$,