If (1+ax)n=1+8x+24x2+…,t then the values of a and n are

Given that, (1+ax)n=1+8x+24x2+… 1+n1ax+n(n−1)1⋅2a2x2+…=1+8x+24x2+…On comparing the coefficients of x,x2 we getna=8,n(n−1)1⋅2a2=24⇒    na(n−1)a=48⇒    8(8−a)=48⇒    8−a=6⇒a=2⇒    n=4