6. Given that the real part of the expansion of $(1+i x)^{n+2}(x \in I)$ is a polynomial in $x$, then the sum of the coefficients of this polynomial is (
(A) $(-1) * 2^{2 \pi+1}$
(B) 0
(C) $2^{2 n+1}$
(D) $-2^{2 x+1}$
6. (B).

Let the real part of the expansion be $f(x)$, and the imaginary part be $g(x)$, then $f(1)$ and $g(1)$ are the sums of the coefficients of the real part polynomial and the imaginary part polynomial, respectively.
$$
\begin{array}{l}
f(1)+i g(1)=(1+i)^{4 n+2} \\
=2^{2 n+1}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{i n+2} \\
=2^{2 n+1}\left[\cos \left(n \pi+\frac{\pi}{2}\right)+i \sin \left(n \pi+\frac{\pi}{2}\right)\right] . \\
=2^{2 n+1} \cdot(-1)^{n} i
\end{array}
$$

It is a pure imaginary number, so $f(1)=0$.