Example 3. If $\mathrm{n}$ is a positive integer, then the value of $\frac{1}{8}\left[1-(-1)^{n}\right]\left(n^{2}-1\right)$ is
(A) definitely odd;
(B) definitely even;
(C) an integer but not necessarily even;
(D) not necessarily an integer.
For the sake of simplicity, let $\left[1-(-1)^{n}\right]\left(n^{2}-1\right)$ $=S$. It is easy to see that, regardless of any positive integer $n$, $8 \mid S$, so $\frac{1}{8} S$ is an integer, thus eliminating (D); it is also evident that when $n$ is odd, $\frac{1}{8} S \neq 0$, so (A) cannot be eliminated; furthermore, when $n$ is even, $\frac{1}{8} S=0$, and when $n$ is odd, we know that the difference between the square of an odd number and 1, $n^{2}-1$, is necessarily a multiple of 8, i.e., $8 \mid \left(n^{2}-1\right)$, and at this time $\left[1-(-1)^{n}\right]=2$, so $\frac{1}{8} S$ is even, thus eliminating (C); and (B) is the correct answer.