Given expansion (1+xn+x253)10 Let x1012=(1)a(xn)b⋅(x253)c Here a,b,c,n are all + ve integers and a≤10, b≤10,c≤4,n≤22,a+b+c=10 Now bn+253c=1012 ⇒bn=253(4−c) For c<4 and n≤22;b>10, which is not possible. ∴c=4,b=0,a=6 ∴x1012=(1)6⋅(xn)0⋅(x253)4 Hence the coefficient of x1012=