2. If $n$ is an integer greater than 1, then the value of $p=n+\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}$
(A) is always even;
(B) is always odd;
(C) is even but not 2;
(D) can be either even or odd.
2. (B)

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
$p=n+\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}$ is always an odd number.