Binomial Theorem
Coefficient in binomial expansion
Grade 11
Question:
<p>If the coefficient of \(x^3\) in the expansion of \((1 + ax + bx^2)(1-2x)^{18}\) is zero, then which of the following pairs \((a, b)\) satisfies this condition?</p>
<p>\(\left(14, \dfrac{272}{3}\right)\)</p>
<p>\(\left(16, \dfrac{272}{3}\right)\)</p>
<p>\(\left(16, \dfrac{272}{5}\right)\)</p>
<p>\(\left(14, \dfrac{272}{5}\right)\)</p>
Step-by-Step Solution
Key Concept: The coefficient of x³ in the expansion comes from three sources: the x³ term from (1-2x)¹⁸, the x² term from (1-2x)¹⁸ multiplied by ax, and the x term from (1-2x)¹⁸ multiplied by bx². Setting their sum to zero gives a linear constraint on a and b.
<p><strong>Step 1:</strong> Identify all sources of x³ term in (1 + ax + bx²)(1-2x)¹⁸</p><p>The x³ coefficient comes from:</p><ul><li>1 × [coefficient of x³ in (1-2x)¹⁸]</li><li>ax × [coefficient of x² in (1-2x)¹⁸]</li><li>bx² × [coefficient of x in (1-2x)¹⁸]</li></ul><p><strong>Step 2:</strong> Calculate coefficients from (1-2x)¹⁸</p><p>• Coefficient of x³: C(18,3)(-2)³ = 816 × (-8) = -6528</p><p>• Coefficient of x²: C(18,2)(-2)² = 153 × 4 = 612</p><p>• Coefficient of x: C(18,1)(-2)¹ = 18 × (-2) = -36</p><p><strong>Step 3:</strong> Set up the equation for x³ coefficient = 0</p><p>-6528 + a(612) + b(-36) = 0</p><p>612a - 36b = 6528</p><p><strong>Step 4:</strong> Simplify by dividing by 36</p><p>17a - b = 181</p><p>∴ The pair (a, b) must satisfy: <strong>b = 17a - 181</strong></p><p>Answer: B (Check which option satisfies this relation)</p>
Correct Answer: B