<p>The coefficient of \(x^5\) in \((1 + 2x + 3x^2 + \cdots)^{3/2}\) is \((|x| < 1)\)</p>
Step-by-Step Solution
Key Concept: Recognize that 1 + 2x + 3x² + ... = d/dx(x + x² + x³ + ...) = d/dx[x/(1-x)] = 1/(1-x)², so we need the coefficient of x⁵ in [1/(1-x)²]^(3/2) = (1-x)^(-3).
<p><strong>Step 1:</strong> Simplify the series: Notice that 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: (1 + 2x + 3x² + ...)^(3/2) = [1/(1-x)²]^(3/2) = (1-x)^(-3)</p><p><strong>Step 3:</strong> Apply generalized binomial theorem: (1-x)^(-3) = Σ C(-3, r)(-x)^r where C(-3, r) = (-3)(-3-1)(-3-2)···(-3-r+1)/r!</p><p><strong>Step 4:</strong> Find coefficient of x⁵ (r = 5): C(-3, 5) = (-3)(-4)(-5)(-6)(-7)/5! = -(3·4·5·6·7)/120 = -2520/120 = -21</p><p><strong>Step 5:</strong> Multiply by (-1)⁵ from (-x)⁵: Coefficient = (-21)(-1) = 21</p><p>∴ Answer: A</p>
Correct Answer: A