If the coefficients of x^2 and x^3 are both zero, in the expansion of the expression (1 + ax + bx^2) (1 - 3x)^15 in powers of x, then the ordered pair (a, b) is equal to:

Correct option is A (28,315)
Step 1:- Find the coefficient of x^2
Given, (1+ax+bx^2)(1
We know that, The coefficient of x^r in the binomial expansion (a+b)^n = nCr * a^(n-r) * b^r
The coefficient of x^r in the binomial expansion (1-3x)^15= 15Cr(-3)^r
The coefficient of x^2 in the binomial expansion (1+ax+bx^2)(1-3x)^15=1× coefficient of x^2 in the binomial expansion (1-3x)^15+a× coefficient of x in the binomial expansion (1-3x)^15+b× coefficient of x^0 in the binomial expansion (1-3x)^15
= 15C2 × (-3)^2 + a × 15C1 × (-3)^1 + b × 15C0 × (-3)^0
=105 × 9 + a × 15 × (-3) + b × 1
=945 - 45a + b
The coefficient of x^2 in the binomial expansion (1+ax+bx^2)(1-3x)^15 = 945 - 45a + b (1)
Step 2:- Find the coefficient of x^3
The coefficient of x^3 in the binomial expansion (1+ax+bx^2)(1-3x)^15=1× coefficient of x^3 in the binomial expansion (1-3x)^15 + a × coefficient of x^2 in the binomial expansion (1-3x)^15 + b × coefficient of x in the binomial expansion (1-3x)^15
= 15C3 × (-3)^3 + a × 15C2 × (-3)^2 + b × 15C1 × (-3)^1
=455 × (-27) + 105 × a × 9 + b × 15 × (-3)
=-12285 + 945a - 45b
The coefficient of x^3 in the binomial expansion (1+ax+bx^2)(1-3x)^15 = -12285 + 945a - 45b (2)
Step 2:- Find the value of a and b
Given, The coefficients of x^2 and x^3 are both zero.
From equation (1) and (2)
945 - 45a + b = 0 (3)
-12285 + 945a - 45b = 0 (4)
From equation (3)
b = 45a - 945
Substitute this value of b in equation (4)
-12285 + 945a - 45(45a - 945) = 0
-12285 + 945a - 2025a + 42525 = 0
-1080a = -30240
a = 28
Substitute a = 28 in b = 45a - 945
b = 45(28) - 945
b = 1260 - 945
b = 315
So, (a,b) = (28,315)
Hence, A is the correct option.