$\begin{array}{l}\text { 6. Given }\left(1+x+x^{2}+x^{3}\right)^{400} \\ =\sum_{k=0}^{300} c_{4 k} x^{4 k}+\sum_{k=0}^{299} c_{4 k+1} x^{4 k+1}+ \\ \sum_{k=0}^{299} c_{4 k+2} x^{4 k+2}+\sum_{k=0}^{299} c_{4 k+3} x^{4 k+3} \text {. } \\ \text { then } \sum_{k=0}^{300} c_{4 k}=\end{array}$
$6.4^{399}$.
$$
\begin{array}{l}
\text { Let } \sum_{k=0}^{300} c_{4 k}=A_{1}, \sum_{k=0}^{299} c_{4 k+1}=A_{2} \text {, } \\
\sum_{k=0}^{299} c_{4 k+2}=A_{3}, \sum_{k=0}^{299} c_{4 k+3}=A_{4} \text {. } \\
\text { Let } x=1, \mathrm{i},-1,-\mathrm{i} \text {, we get } \\
\left\{\begin{array}{l}
4^{400}=A_{1}+A_{2}+A_{3}+A_{4}, \\
0=A_{1}+\mathrm{i} A_{2}-A_{3}-\mathrm{i} A_{4}, \\
0=A_{1}-A_{2}+A_{3}-A_{4}, \\
0=A_{1}-\mathrm{i} A_{2}-A_{3}+\mathrm{i} A_{4} .
\end{array}\right. \\
\end{array}
$$

Solving, we get $A_{1}=A_{2}=A_{3}=A_{4}=4^{399}$.