$11 \cdot 2$ In the expansion of $\left(a-\frac{1}{\sqrt{a}}\right)^{7}$, the coefficient of $a^{-\frac{1}{2}}$ is
(A) -7 .
(B) 7 .
(C) -21 .
(D) 21 .
(E) 35 .
(14th American High School Mathematics Examination, 1963)
[Solution] By the binomial theorem, the $(r+1)$-th term of the above expansion is
$$
C_{7}^{r} a^{7-r}(-1)^{r}\left(a^{-\frac{1}{2}}\right)^{r}=(-1)^{r} C_{7}^{r} \cdot a^{7-\frac{3}{2} r} \text {, }
$$

Given that $\quad 7-\frac{3}{2} r=-\frac{1}{2}$, we get $r=5$. Therefore, the coefficient of $a^{-\frac{1}{2}}$ is $\quad(-1)^{5} C_{7}^{5}=-C_{7}^{2}=-21$. Hence, the answer is (C).