Consider the term, (1+x)−2=1−2x+3x2−4x3+.... Simplify the above term as, (1−2x+3x2−4x3+....)−n=[(1+x)−2]−n =(1+x)2n General term in the expansion of (1+x)2n is given as, Tr+1=2nCrxr T6+1=2nC6x6 The coefficient of x6 for the given expansion is ‌2nC6