$30 \cdot 3$ If $n$ is an integer greater than 1, then the value of the algebraic expression
$$
p=n+\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}
$$

is
(A) always even.
(B) always odd.
(C) even but not 2.
(D) can be either even or odd.
(China Junior High School Mathematics League, 1985)
[Solution] When $n$ is even, $\frac{1-(-1)^{n}}{2}=0, p=n+1$ is odd; when $n$ is odd, $\frac{1-(-1)^{n}}{2}=1$, and $n^{2}-1$ is even, so $p=n+\left(n^{2}-1\right)$ is odd. Therefore, the answer is $(B)$.