The number of real negative terms in ${(1+ix)}^{4n-2},n\in N,x>0$ is 

${(1+ix)}^{4n-2}$ total number of terms $=4n-1$

${(1+ix)}^{4n-2}={}^{4n-2}{\text{C}}_0+{}^{4n-2}{\text{C}}_1\left(ix\right)+{}^{4n-2}{\text{C}}_2{\left(ix\right)}^2+\cdots$

$=\text{Real}+\text{Complex}+\text{Real}+\text{Complex}+\cdots$

In $4n-1$ terms 2n terms are real & $2n-1$ are complex

In 2n terms n are real positive and n are real negative