Binomial Theorem
Coefficient in infinite series expansion
Grade 11
Question:
<p>In the expansion of \((1 + 2x + 3x^2 + \cdots)^{-3/2}\), there is no term containing \(x^5\). Is this statement true?</p>
<p>False</p>
<p>True</p>
<p>Cannot be determined</p>
<p>Depends on the range of x</p>
Step-by-Step Solution
Key Concept: Recognize that (1 + 2x + 3x² + ...) = 1/(1-x)², so the expression becomes (1-x)³. The expansion of (1-x)³ is finite with only 4 terms, making the coefficient of x⁵ automatically zero.
<p><strong>Step 1:</strong> Recognize the series in the base: 1 + 2x + 3x² + 4x³ + ... = d/dx[x + x² + x³ + ...] = d/dx[x/(1-x)] = 1/(1-x)²</p><p><strong>Step 2:</strong> Rewrite the expression as: [1/(1-x)²]^(-3/2) = (1-x)³</p><p><strong>Step 3:</strong> Expand using binomial theorem: (1-x)³ = 1 - 3x + 3x² - x³</p><p><strong>Step 4:</strong> This polynomial has only 4 terms (up to x³). There is no x⁴ or x⁵ term in the expansion.</p><p>∴ The statement is <strong>TRUE</strong> — there is no term containing x⁵.</p>
Correct Answer: B