If the coefficients of x2 and x3 are both zero, in the expansion of the expression1+ax+bx2(1−3x)15 in powers of x then the ordered pair (a, b) is equal to

Given expression is 1+ax+bx2(1−3x)15 in the expansion of binomial (1−3x)15 the (r+1)th term is Tr+1=15Cr(−3x)r=15Cr(−3)rxrNow, coefficient of x2, in the expansion of1+ax+bx2(1−3x)15 15C2(−3)2+a15C1(−3)1+b15C0(−3)0=0  (given)⇒ (105×9)−45a+b=0⇒45a−b=945  …..(1)similarly, the" coefficient of x3, in the expansion of1+ax+bx2(1−3x)15 is  15C3(−3)3+a15C2(−3)2+b15C1(−3)1=0⇒    −12285+945a−45b    =0⇒        63a−3b=819⇒21a−b    =273----iFrom Eqs. (i) and (ii), we get 24a=672⇒a=28so, b=315⇒(a,b)=(28,315)