$1 x \in \mathbf{R},(1+i x)^{50}$ In the expansion, the sum of the coefficients of the real terms is
A. $2^{25}$
B. $-2^{25}$
C. 0
D. Not the above answers
1 C. The real coefficient terms are the even powers of $x$. The sum of the required coefficients is
$$
\frac{1}{2}\left((1+i)^{50}+(1-i)^{50}\right)=\frac{1}{2}\left((2 i)^{25}+(-2 i)^{25}\right)=0 .
$$