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

Step 1: Understanding the Concept:
We need to expand the product and isolate the coefficients of \(x^{3}\) and \(x^{4}\). Each coefficient will involve \(a\) and \(b\). Setting them to zero creates a system of linear equations.
Step 3: Detailed Explanation:
1. General term of \((1 - 2x)^{18}\) is \(C_{r} = \binom{18}{r}(-2x)^{r}\).
2. Coefficient of \(x^{3}\) in \((1 + ax + bx^{2})(1 - 2x)^{18}\):
\[ \text{Coeff}(x^{3}) = 1 \cdot \binom{18}{3}(-2)^{3} + a \cdot \binom{18}{2}(-2)^{2} + b \cdot \binom{18}{1}(-2)^{1} = 0 \]
\[ -8(816) + 4a(153) - 2b(18) = 0 \implies -6528 + 612a - 36b = 0 \]
\[ 17a - b = \frac{6528}{36} = \frac{544}{3} \dots (i) \]
3. Coefficient of \(x^{4}\) in \((1 + ax + bx^{2})(1 - 2x)^{18}\):
\[ \text{Coeff}(x^{4}) = 1 \cdot \binom{18}{4}(-2)^{4} + a \cdot \binom{18}{3}(-2)^{3} + b \cdot \binom{18}{2}(-2)^{2} = 0 \]
\[ 16(3060) - 8a(816) + 4b(153) = 0 \implies 48960 - 6528a + 612b = 0 \]
\[ 32a - 3b = \frac{48960}{204} = 240 \dots (ii) \]
4. Solving equations (i) and (ii):
Multiply (i) by 3: \(51a - 3b = 544\).
Subtract (ii) from this: \((51 - 32)a = 544 - 240 \implies 19a = 304 \implies a = 16\).
Substitute \(a=16\) in (ii): \(3b = 32(16) - 240 = 512 - 240 = 272 \implies b = 272/3\).
Step 4: Final Answer:
The values are \((16, \frac{272}{3})\).