$7 \cdot 104$ If $f(a+b)=f(a) \cdot f(b)$, and $f(1)=1$. Then
$$
\frac{f(2)}{f(1)}+\frac{f(3)}{f(2)}+\frac{f(4)}{f(3)}+\cdots+\frac{f(1988)}{f(1987)}
$$

is
(A) 1986.
(B) 1987.
(C) 1988.
(D) Cannot be determined.
(Suzhou, Jiangsu Province, China High School Mathematics Competition, 1988)
[Solution] Let $a=1, b=1$, then $f(2)=f(1) \cdot f(1)$, and $f(1)=1$,
$$
\therefore \quad \frac{f(2)}{f(1)}=f(1)=1 \text {. }
$$

Let $a=2, b=1$, then $f(3)=f(2) \cdot f(1)$,
$$
\therefore \quad \frac{f(3)}{f(2)}=f(1)=1 \text {. }
$$

By induction, we have $\quad \frac{f(1988)}{f(1987)}=f(1)=1$,
$$
\therefore \quad \text { the original expression }=1987 .
$$

Therefore, the answer is $(B)$.