7. Let $f(x)=a x+b(a, b$ be real numbers),
$$
\begin{array}{l}
f_{1}(x)=f(x), \\
f_{n+1}(x)=f\left(f_{n}(x)\right)(n=1,2, \cdots) .
\end{array}
$$

If $f_{7}(x)=128 x+381$, then $a+b=$
Ni.7.5.
From the problem, we know
$$
\begin{array}{l}
f_{n}(x)=a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
=a^{n} x+\frac{a^{n}-1}{a-1} \cdot b .
\end{array}
$$

From $f_{7}(x)=128 x+381$, we get
$$
a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381 \text {. }
$$

Therefore, $a=2, b=3, a+b=5$.