Question

If the coefficients of \(x^3\) and \(x^4\) in the expansion of \((1 + ax + bx^2)(1 - 2x)^{18}\) are both zero, then \((a,b)\) is equal to

• A. \( \left(14, \frac{272}{3}\right) \)
• B. \( \left(16, \frac{272}{3}\right) \)
• C. \( \left(16, \frac{251}{3}\right) \)
• D. \( \left(14, \frac{251}{3}\right) \)

Hint:

Break coefficient questions into contributions from each term — systematic approach avoids mistakes.

Answer 1

The Correct Option is B

Concept: General term of \((1 - 2x)^{18}\): \[ T_r = \binom{18}{r} (-2x)^r \]

Step 1: Coefficient of \(x^3\)
From: \[ (1)(x^3 \text{ term}) + (ax)(x^2 \text{ term}) + (bx^2)(x \text{ term}) \] \[ = \binom{18}{3}(-2)^3 + a\binom{18}{2}(-2)^2 + b\binom{18}{1}(-2) \] \[ = -8\binom{18}{3} + 4a\binom{18}{2} - 2b\binom{18}{1} \] Set = 0

Step 2: Coefficient of \(x^4\)
\[ = \binom{18}{4}(-2)^4 + a\binom{18}{3}(-2)^3 + b\binom{18}{2}(-2)^2 \] \[ = 16\binom{18}{4} - 8a\binom{18}{3} + 4b\binom{18}{2} \] Set = 0

Step 3: Substitute values \[ \binom{18}{1}=18,\quad \binom{18}{2}=153,\quad \binom{18}{3}=816,\quad \binom{18}{4}=3060 \] First equation: \[ -8(816) + 4a(153) - 2b(18) = 0 \] \[ -6528 + 612a - 36b = 0 \quad ...(1) \] Second equation: \[ 16(3060) - 8a(816) + 4b(153) = 0 \] \[ 48960 - 6528a + 612b = 0 \quad ...(2) \]

Step 4: Solve equations From (1): \[ 612a - 36b = 6528 \Rightarrow 17a - b = 181.33 \quad ...(3) \] Solving with (2), we get: \[ a = 16,\quad b = \frac{272}{3} \] \[ \therefore (a,b) = \left(16, \frac{272}{3}\right) \]