11.18 The expression $\frac{(1+\sqrt{1991})^{100}-(1-\sqrt{1991})^{100}}{\sqrt{1991}}$ represents a number that is
(A) a natural number.
(B) an integer but not a natural number.
(C) a rational number but not an integer.
(D) an irrational number.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
［Solution］Notice that
$$
\begin{aligned}
& (1+\sqrt{1991})^{100}-(1-\sqrt{1991})^{100} \\
= & 2\left[C_{100}^{1}(\sqrt{1991})^{1}+C_{100}^{3}(\sqrt{1991})^{3}+\cdots+C_{100}^{99}(\sqrt{1991})^{99}\right],
\end{aligned}
$$

then
$$
\begin{aligned}
\text { Original expression } & =2\left[C_{100}^{1}+C_{100}^{3}(\sqrt{1991})^{2}+\cdots+C_{100}^{99}(\sqrt{1991})^{98}\right] \\
& =2\left(C_{100}^{1}+C_{100}^{3} \cdot 1991+\cdots+C_{100}^{99} \cdot 1991^{49}\right),
\end{aligned}
$$

it is a positive integer,
thus the answer is $(A)$.