7. Let $f(x)=a x+b$, where $a, b$ are real numbers, $f_{1}(x)=f(x), f_{n+1}(x)=f\left(f_{n}(x)\right)$, $n=1,2, \cdots$, If $f_{7}(x)=128 x+381$, then $a+b=$ $\qquad$ .
$$
\begin{array}{l}
f_{2}(x)=a(a x+b)+b=a^{2} x+a b+b=a^{2} x+\frac{b\left(1-a^{2}\right)}{1-a}, \\
f_{3}(x)=a\left(a^{2} x+a b+b\right)+b=a^{3} x+a^{2} b+a b+b=a^{3} x+\frac{b\left(1-a^{3}\right)}{1-a}, \cdots, \\
f_{7}(x)=a^{7} x+\frac{b\left(1-a^{7}\right)}{1-a}=128 x+381 \Rightarrow a=2, b=3 . \text{ So } a+b=5 .
\end{array}
$$