The number of real negative terms in the binomial expansion of (1+ix)4n−2,n∈N,x>0 is

Tr+1=4n−2Cr(ix)rTr+1 is negative, if ir is negative and real. ir=−1⇒ r=2,6,10,… which form an A.P.with common difference 4 0≤r≤4n−24n−2=2+(r−1)4⇒ r=nThe required number of terms is n.