1. Given that the real part of $(1+\mathrm{i} x)^{4 n+2}(x \in \mathbf{R})$ is a polynomial in $x$, then the sum of the coefficients of this polynomial is $\qquad$ .
1. 0 Explanation: Let the real and imaginary parts of the expansion be $f(x)$ and $g(x)$, respectively. Then $f(1)$ and $g(1)$ are the sums of the coefficients of the real and imaginary part polynomials, $\therefore f(1)+\mathrm{i} g(1)=(1+\mathrm{i})^{4 n+2}=2^{2 n+1}\left(\cos \frac{\pi}{4}+\mathrm{i} \sin \frac{\pi}{4}\right)^{4 n+2}$ $=2^{2 n+1}\left[\cos \left(n \pi+\frac{\pi}{2}\right)+\mathrm{i} \sin \left(n \pi+\frac{\pi}{2}\right)\right]=2^{2 n+1} \cdot(-1)^{n} \mathrm{i}$ is a pure imaginary number, $\therefore f(1)=0$.