China Mathematical Competition

$$\begin{array}{rl}{f}_n(x)& =a^{n}x+(a^{n-1}+a^{n-2}+\cdots +a+1)b\\ & =a^{n}x+\frac{a^n-1}{a-1}\times b.\end{array}$$ As $f_7(x)=128x+381$, we have $a^7=128$ and $\frac{a^7-1}{a-1}\times b=381$. Then $a=2$, $b=3$. The answer is $a+b=5$.