Given $\left.(1+\mathrm{i} r)^{2}\right)^{201}=f(x) \div \mathrm{i} g(x)$, where $f, g$ are polynomials with real coefficients, then the sum of the coefficients of $f(x)$ is
(A) 0
(B) $2^{\text {no }}$
(C) $2^{1000}$
(D) $2^{20017}$
5. (C).

Given the condition $f(1)+\mathrm{i} g(1) \cdots(1+\mathrm{i})^{2001}=(2 \mathrm{i})^{1000}(1+\mathrm{i})=2^{1000}+2^{1000} \mathrm{i}$, so the sum of the coefficients of $f(x)$, $f(1)=2^{1000}$.