On replacing $x$ by $-\frac{1}{x}$, we get
$$
\begin{aligned}
& \frac{\left(1-x+x^2\right)^n}{x^{2 n}} \\
& =\frac{C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}-C_3 x^{2 n-3}+\ldots}{x^{2 n}} \\
& \Rightarrow\left(1-x+x^2\right)^n=C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}
\end{aligned}
$$

$$
\begin{aligned}
& \because C_0 C_1-C_1 C_2+C_2 C_3-\ldots=\text { Coefficient of } x^{2 n+1} \text { in } \\
& \left(C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}-C_3 x^{2 n-3}+\ldots\right) \\
& \times\left(C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots\right) \\
& =\text { Coefficient of } x^{2 n+1} \text { in }\left(1+x+x^2\right)^n\left(1-x+x^2\right)^n \\
& =\text { Coefficient of } x^{2 n+1} \text { in }\left(\left(1+x^2\right)^2-x^2\right)^n \\
& =\text { Coefficient of } x^{2 n+1} \text { in }\left(1+x^4+x^2\right)^n=0
\end{aligned}
$$
$\because$ Coefficient of $x^k$ in the expansion of $\left(1+x^2+x^4\right)^n$ exists if $k$ is even, otherwise it will be zero and $(2 n+1)$ is odd.

Hence, option (b) is correct.