3. If $n$ is an integer greater than 1, then the value of $p=n+$ $\left(n^{2}-1\right)^{\frac{1-(-1)^{2}}{2}}$ is ( ).
(A) always even
(B) always odd
(C) even, but not 2
(D) can be either even or odd
3. Solve for $n$ being even, $\frac{i-(-1)^{n}}{2}=0$. $\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}=1$, then $p=n+1$ is odd;
$n$ being odd, $\frac{1-(-1)^{n}}{2}=1$, $\left(n^{2}-1\right)^{\frac{1-(-1)^{n}}{2}}=n^{2}-1$, then $p=n+n^{2}-1$ is odd. Choose B.